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The simplified value of following is: $(\frac{3}{15} a^{5} b^{6} c^{3} \times \frac{5}{9} a b^{5} c^{4})$ $\left( {\frac{3}{{15}}{a^5}{b^6}{c^3} \times \frac{5}{9}a{b^5}{c^4}} \right)$ ÷ $\frac{10}{17} a^{2} b c^{3}$ $\frac{{10}}{{17}}{a^2}b{c^3}$

1 $\frac{3}{10} a b^{4} c^{3}$ $\frac{3}{{10}}a{b^4}{c^3}$
2 $\frac{9}{10} a^{2} b c^{4}$ $\frac{9}{{10}}{a^2}b{c^4}$
3 $\frac{9}{10} a^{2} b c^{4}$ $\frac{9}{{10}}{a^2}b{c^4}$
4 $\frac{1}{10} a^{4} b^{4} c^{10}$ $\frac{1}{{10}}{a^4}{b^4}{c^{10}}$

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} (\frac{3}{15} a^{5} b^{6} c^{3} \times \frac{5}{9} a b^{5} c^{4}) \div \frac{10}{27} a^{2} b c^{3} \\ \Rightarrow \frac{1}{9} a^{6} b^{11} c^{7} \div \frac{10}{27} a^{2} b c^{3} \\ \Rightarrow \frac{\frac{1}{9} a^{6} b^{11} c^{7}}{\frac{10}{27} a^{2} b c^{3}} \\ \Rightarrow \frac{3}{10} a^{4} b^{10} c^{4} \end{aligned}

$\eqalign{ & \left( {\frac{3}{{15}}{a^5}{b^6}{c^3} \times \frac{5}{9}a{b^5}{c^4}} \right) \div \frac{{10}}{{27}}{a^2}b{c^3} \cr & \Rightarrow \frac{1}{9}{a^6}{b^{11}}{c^7} \div \frac{{10}}{{27}}{a^2}b{c^3} \cr & \Rightarrow \frac{{\frac{1}{9}{a^6}{b^{11}}{c^7}}}{{\frac{{10}}{{27}}{a^2}b{c^3}}} \cr & \Rightarrow \frac{3}{{10}}{a^4}{b^{10}}{c^4} \cr}$

02. If a² + b² + c² = 2(a - b - c) -3, then the value of a + b + c is?

1-2
21
32
4-1

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} a^{2} + b^{2} + c^{2} = 2 (a - b - c) - 3 \\ \Rightarrow a^{2} + b^{2} + c^{2} = 2 a - 2 b - 2 c - 3 \\ \Rightarrow a^{2} + b^{2} + c^{2} - 2 a + 2 b + 2 c + 3 = 0 \\ \Rightarrow a^{2} - 2 a + 1 + b^{2} + 2 b + 1 + c^{2} + 2 c + 1 = 0 \\ \Rightarrow {(a - 1)}^{2} + {(b + 1)}^{2} + {(c + 1)}^{2} = 0 \\ {(a - 1)}^{2} = 0 {(b + 1)}^{2} = 0 {(c + 1)}^{2} = 0 \\ \Rightarrow a - 1 = 0 \Rightarrow b + 1 = 0 \Rightarrow c + 1 = 0 \\ \Rightarrow a = 1 \Rightarrow b = - 1 \Rightarrow c = - 1 \\ ∴ a + b + c \\ = 1 - 1 - 1 \\ = - 1 \end{aligned}

$\eqalign{ & {a^2} + {b^2} + {c^2} = 2\left( {a - b - c} \right) - 3 \cr & \Rightarrow {a^2} + {b^2} + {c^2} = 2a - 2b - 2c - 3 \cr & \Rightarrow {a^2} + {b^2} + {c^2} - 2a + 2b + 2c + 3 = 0 \cr & \Rightarrow {a^2} - 2a + 1 + {b^2} + 2b + 1 + {c^2} + 2c + 1 = 0 \cr & \Rightarrow {\left( {a - 1} \right)^2} + {\left( {b + 1} \right)^2} + {\left( {c + 1} \right)^2} = 0 \cr & {\left( {a - 1} \right)^2} = 0{\text{ }}{\left( {b + 1} \right)^2} = 0{\text{ }}{\left( {c + 1} \right)^2} = 0 \cr & \Rightarrow a - 1 = 0{\text{ }} \Rightarrow b + 1 = 0{\text{ }} \Rightarrow c + 1 = 0 \cr & \Rightarrow a = 1{\text{ }} \Rightarrow b = - 1{\text{ }} \Rightarrow c = - 1 \cr & \therefore a + b + c \cr & = 1 - 1 - 1 \cr & = - 1 \cr}$

03. If x(x + y + z) = 20, y = (x + y + z) = 30 & z(x + y + z) = 50, then the value of 2(x + y + z) is?

120
210
315
418

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Put (x + y + z) = 10 \\ x = 2 \\ y = 3 \\ z = 5 \\ x (x + y + z) = 20 \\ \Leftrightarrow 2 (10) = 20 \\ \Leftrightarrow 20 = 20 \\ Similarly other will satisfied, \\ So, value of 2 (x + y + z) \\ = 2 (10) \\ = 20 \end{aligned}

$\eqalign{ & {\text{Put }}\left( {x + y + z} \right) = 10 \cr & x = 2 \cr & y = 3 \cr & z = 5 \cr & x\left( {x + y + z} \right) = 20 \cr & \Leftrightarrow 2\left( {10} \right) = 20 \cr & \Leftrightarrow 20 = 20 \cr & {\text{Similarly other will satisfied,}} \cr & {\text{So, value of 2}}\left( {x + y + z} \right) \cr & = 2\left( {10} \right) \cr & = 20 \cr}$

If x = y = z, then $\frac{{(x + y + z)}^{2}}{x^{2} + y^{2} + z^{2}}$ $\frac{{{{\left( {x + y + z} \right)}^2}}}{{{x^2} + {y^2} + {z^2}}}$ is?

12
23
31
44

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{{(x + y + z)}^{2}}{x^{2} + y^{2} + z^{2}} \\ Assume x = y = z = 1 \\ \Leftrightarrow \frac{{(1 + 1 + 1)}^{2}}{1 + 1 + 1} \\ \Leftrightarrow \frac{9}{3} \\ \Leftrightarrow 3 \end{aligned}

$\eqalign{ & \frac{{{{\left( {x + y + z} \right)}^2}}}{{{x^2} + {y^2} + {z^2}}} \cr & {\text{Assume }}x = y = z = 1 \cr & \Leftrightarrow \frac{{{{\left( {1 + 1 + 1} \right)}^2}}}{{1 + 1 + 1}} \cr & \Leftrightarrow \frac{9}{3} \cr & \Leftrightarrow 3 \cr}$

If $\frac{x}{y} = \frac{a + 2}{a - 2},$ $\frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}}{\text{,}}$ then the value of $\frac{x^{2} - y^{2}}{x^{2} + y^{2}}$ $\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}$ = ?

1 $\frac{2 a}{a^{2} + 2}$ $\frac{{2a}}{{{a^2} + 2}}$
2 $\frac{4 a}{a^{2} + 4}$ $\frac{{4a}}{{{a^2} + 4}}$
3 $\frac{2 a}{a^{2} + 4}$ $\frac{{2a}}{{{a^2} + 4}}$
4 $\frac{4 a}{a^{2} + 2}$ $\frac{{4a}}{{{a^2} + 2}}$

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{x}{y} = \frac{a + 2}{a - 2} \\ \frac{x^{2}}{y^{2}} = \frac{{(a + 2)}^{2}}{{(a - 2)}^{2}} \end{aligned}

$\eqalign{ & \frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}} \cr & \frac{{{x^2}}}{{{y^2}}}{\text{ = }}\frac{{{{\left( {a + 2} \right)}^2}}}{{{{\left( {a - 2} \right)}^2}}} \cr}$
Applying componendo and dividendo

\begin{aligned} ∴ \frac{x^{2} - y^{2}}{x^{2} + y^{2}} \\ = \frac{{(a + 2)}^{2} - {(a - 2)}^{2}}{{(a + 2)}^{2} + {(a - 2)}^{2}} \\ = \frac{8 a}{2 a^{2} + 8} \\ = \frac{4 a}{a^{2} + 4} \end{aligned}

$\eqalign{ & \therefore \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} \cr & = \frac{{{{\left( {a + 2} \right)}^2} - {{\left( {a - 2} \right)}^2}}}{{{{\left( {a + 2} \right)}^2} + {{\left( {a - 2} \right)}^2}}} \cr & = \frac{{8a}}{{2{a^2} + 8}} \cr & = \frac{{4a}}{{{a^2} + 4}} \cr}$

If $x - \frac{1}{x} = 2,$ $x - \frac{1}{x} = 2{\text{,}}$ then the value of the following is $x^{3} - \frac{1}{x^{3}}$ ${x^3} - \frac{1}{{{x^3}}}$ = ?

12
211
315
414

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} x - \frac{1}{x} = 2 to find x^{3} - \frac{1}{x^{3}} \\ \Rightarrow x - \frac{1}{x} = 2 \\ [Cubing both sides] \\ \Rightarrow {(x - \frac{1}{x})}^{3} = {(2)}^{3} \\ \Rightarrow x^{3} - \frac{1}{x^{3}} - 3 \times x \times \frac{1}{x} (x - \frac{1}{x}) = 8 \\ \Rightarrow x^{3} - \frac{1}{x^{3}} - 3 \times (2) = 8 \\ \Rightarrow x^{3} - \frac{1}{x^{3}} = 14 \end{aligned}

$\eqalign{ & x - \frac{1}{x} = 2{\text{ to find }}{x^3} - \frac{1}{{{x^3}}} \cr & \Rightarrow x - \frac{1}{x} = 2 \cr & \left[ {{\text{Cubing both sides}}} \right] \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^3} = {\left( 2 \right)^3} \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3 \times x \times \frac{1}{x}\left( {x - \frac{1}{x}} \right) = 8 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3 \times \left( 2 \right) = 8 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 14 \cr}$

07. If a² + b² + c² - ab - bc - ca = 0 then a : b : c is?

11 : 2 : 1
22 : 1 : 1
31 : 1 : 2
41 : 1 : 1

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given, \\ a^{2} + b^{2} + c^{2} - a b - b c - c a = 0 \\ Find, a : b : c = ? \\ According to the question, \\ a^{2} + b^{2} + c^{2} - a b - b c - c a = 0 \\ \Rightarrow a^{2} + b^{2} + c^{2} = a b + b c + c a \\ Let a = b = c = 1 \\ \Rightarrow 1^{2} + 1^{2} + 1^{2} = 1 \times 1 + (1 \times 1) + (1 \times 1) \\ \Rightarrow 3 = 3 \\ \Rightarrow So, ratio of a : b : c = 1 : 1 : 1 \end{aligned}

$\eqalign{ & {\text{Given, }} \cr & {a^2} + {b^2} + {c^2} - ab - bc - ca = 0 \cr & {\text{Find, }}a:b:c = ? \cr & {\text{According to the question, }} \cr & {a^2} + {b^2} + {c^2} - ab - bc - ca = 0 \cr & \Rightarrow {a^2} + {b^2} + {c^2} = ab + bc + ca \cr & {\text{Let }}a = b = c = 1 \cr & \Rightarrow {1^2} + {1^2} + {1^2} = 1 \times 1 + \left( {1 \times 1} \right) + \left( {1 \times 1} \right) \cr & \Rightarrow 3 = 3 \cr & \Rightarrow {\text{So, ratio of }}a:b:c = 1:1:1 \cr}$

If $\frac{x + 1}{x - 1} = \frac{a}{b}$ $\frac{{x + 1}}{{x - 1}} = \frac{a}{b}$ and $\frac{1 - y}{1 + y} = \frac{b}{a},$ $\frac{{1 - y}}{{1 + y}} = \frac{b}{a}{\text{,}}$ then the value of $\frac{x - y}{1 + x y}$ $\frac{{x - y}}{{1 + xy}}$ is?

1 $\frac{a^{2} - b^{2}}{a b}$ $\frac{{{a^2} - {b^2}}}{{ab}}$
2 $\frac{a^{2} + b^{2}}{2 a b}$ $\frac{{{a^2} + {b^2}}}{{2ab}}$
3 $\frac{a^{2} - b^{2}}{2 a b}$ $\frac{{{a^2} - {b^2}}}{{2ab}}$
4 $\frac{2 a b}{a^{2} - b^{2}}$ $\frac{{2ab}}{{{a^2} - {b^2}}}$

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given , \\ \frac{x + 1}{x - 1} = \frac{a}{b} \\ (Using componendo &  dividendo) \\ \Leftrightarrow \frac{x}{1} = \frac{a + b}{a - b} \\ \Leftrightarrow x = \frac{a + b}{a - b} . . . . . (i) \\ Again, \frac{1 - y}{1 + y} = \frac{b}{a} \\ \Leftrightarrow \frac{1 + y}{1 - y} = \frac{a}{b} \\ \Leftrightarrow \frac{1}{y} = \frac{a + b}{a - b} \\ \Leftrightarrow y = \frac{a - b}{a + b} . . . . . (i i) \\ From question, \\ \frac{x - y}{1 + x y} \\ \Rightarrow \frac{\frac{a + b}{a - b} - \frac{a - b}{a + b}}{1 + (\frac{a + b}{a - b}) (\frac{a - b}{a + b})} \\ \Rightarrow \frac{{(a + b)}^{2} - {(a - b)}^{2}}{(a^{2} - b^{2}) (1 + 1)} \\ \Rightarrow \frac{4 a b}{2 (a^{2} - b^{2})} \\ \Rightarrow \frac{2 b}{a^{2} - b^{2}} \end{aligned}

$\eqalign{ & {\text{Given ,}} \cr & \frac{{x + 1}}{{x - 1}} = \frac{a}{b} \cr & \left( {{\text{Using componendo & dividendo}}} \right) \cr & \Leftrightarrow \frac{x}{1} = \frac{{a + b}}{{a - b}} \cr & \Leftrightarrow x = \frac{{a + b}}{{a - b}}\,.....(i) \cr & {\text{Again,}}\frac{{1 - y}}{{1 + y}} = \frac{b}{a} \cr & \Leftrightarrow \frac{{1 + y}}{{1 - y}} = \frac{a}{b} \cr & \Leftrightarrow \frac{1}{y} = \frac{{a + b}}{{a - b}} \cr & \Leftrightarrow y = \frac{{a - b}}{{a + b}}\,.....(ii) \cr & {\text{From question,}} \cr & \frac{{x - y}}{{1 + xy}} \cr & \Rightarrow \frac{{\frac{{a + b}}{{a - b}} - \frac{{a - b}}{{a + b}}}}{{1 + \left( {\frac{{a + b}}{{a - b}}} \right)\left( {\frac{{a - b}}{{a + b}}} \right)}} \cr & \Rightarrow \frac{{{{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2}}}{{\left( {{a^2} - {b^2}} \right)\left( {1 + 1} \right)}} \cr & \Rightarrow \frac{{4ab}}{{2\left( {{a^2} - {b^2}} \right)}} \cr & \Rightarrow \frac{{2b}}{{{a^2} - {b^2}}} \cr}$

09. If x + y + z = 6 and xy + yz + zx = 10, then the value of x³ + y³ + z³ - 3xyz is?

136
240
342
448

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given \\ x + y + z = 6 \\ x y + y z + z x = 10 \\ To find x^{3} + y^{3} + z^{3} - 3 x y z =  ? \\ \Rightarrow Using formula, \\ \Rightarrow {(x + y + z)}^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + y z + z x) \\ \Rightarrow 6^{2} = x^{2} + y^{2} + z^{2} + 2 \times 10 \\ \Rightarrow 36 = x^{2} + y^{2} + z^{2} + 20 \\ \Rightarrow x^{2} + y^{2} + z^{2} = 16 \\ \Rightarrow x^{2} + y^{2} + z^{2} - 3 x y z = (x + y + z) (x^{2} + y^{2} + z^{2} - x y - y z - z x) \\ \Rightarrow x^{2} + y^{2} + z^{2} - 3 x y z = 6 [16 - 10] \\ \Rightarrow x^{2} + y^{2} + z^{2} - 3 x y z = 6 \times 6 \\ \Rightarrow x^{2} + y^{2} + z^{2} - 3 x y z = 36 \end{aligned}

$\eqalign{ & {\text{Given}} \cr & x + y + z = 6 \cr & xy + yz + zx = 10 \cr & {\text{To find }}{x^3} + {y^3} + {z^3} - 3xyz{\text{ = ?}} \cr & \Rightarrow {\text{ Using formula,}} \cr & \Rightarrow {\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + zx} \right) \cr & \Rightarrow {6^2} = {x^2} + {y^2} + {z^2} + 2 \times 10 \cr & \Rightarrow 36 = {x^2} + {y^2} + {z^2} + 20 \cr & \Rightarrow {x^2} + {y^2} + {z^2} = 16 \cr & \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right) \cr & \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = 6\left[ {16 - 10} \right] \cr & \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = 6 \times 6 \cr & \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = 36 \cr}$

The sum of $\frac{1}{x + y}$ $\frac{1}{{x + y}}$ and $\frac{1}{x - y}$ $\frac{1}{{x - y}}$ is?

1 $\frac{2 y}{x^{2} - y^{2}}$ $\frac{{2y}}{{{x^2} - {y^2}}}$
2 $\frac{2 x}{x^{2} - y^{2}}$ $\frac{{2x}}{{{x^2} - {y^2}}}$
3 $\frac{- 2 y}{x^{2} - y^{2}}$ $\frac{{ - 2y}}{{{x^2} - {y^2}}}$
4 $\frac{2 x}{y^{2} - x^{2}}$ $\frac{{2x}}{{{y^2} - {x^2}}}$

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} According to the question, \\ \frac{1}{x + y} + \frac{1}{x - y} \\ = \frac{x - y + x + y}{x^{2} - y^{2}} \\ = \frac{2 x}{x^{2} - y^{2}} \end{aligned}

$\eqalign{ & {\text{According to the question,}} \cr & \frac{1}{{x + y}} + \frac{1}{{x - y}} \cr & = \frac{{x - y + x + y}}{{{x^2} - {y^2}}} \cr & = \frac{{2x}}{{{x^2} - {y^2}}} \cr}$

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algebra - 05

The simplified value of following is: $(\frac{3}{15} a^{5} b^{6} c^{3} \times \frac{5}{9} a b^{5} c^{4})$ $\left( {\frac{3}{{15}}{a^5}{b^6}{c^3} \times \frac{5}{9}a{b^5}{c^4}} \right)$ ÷ $\frac{10}{17} a^{2} b c^{3}$ $\frac{{10}}{{17}}{a^2}b{c^3}$

02. If a² + b² + c² = 2(a - b - c) -3, then the value of a + b + c is?

03. If x(x + y + z) = 20, y = (x + y + z) = 30 & z(x + y + z) = 50, then the value of 2(x + y + z) is?

If x = y = z, then $\frac{{(x + y + z)}^{2}}{x^{2} + y^{2} + z^{2}}$ $\frac{{{{\left( {x + y + z} \right)}^2}}}{{{x^2} + {y^2} + {z^2}}}$ is?

If $\frac{x}{y} = \frac{a + 2}{a - 2},$ $\frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}}{\text{,}}$ then the value of $\frac{x^{2} - y^{2}}{x^{2} + y^{2}}$ $\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}$ = ?

If $x - \frac{1}{x} = 2,$ $x - \frac{1}{x} = 2{\text{,}}$ then the value of the following is $x^{3} - \frac{1}{x^{3}}$ ${x^3} - \frac{1}{{{x^3}}}$ = ?

07. If a² + b² + c² - ab - bc - ca = 0 then a : b : c is?

If $\frac{x + 1}{x - 1} = \frac{a}{b}$ $\frac{{x + 1}}{{x - 1}} = \frac{a}{b}$ and $\frac{1 - y}{1 + y} = \frac{b}{a},$ $\frac{{1 - y}}{{1 + y}} = \frac{b}{a}{\text{,}}$ then the value of $\frac{x - y}{1 + x y}$ $\frac{{x - y}}{{1 + xy}}$ is?

09. If x + y + z = 6 and xy + yz + zx = 10, then the value of x³ + y³ + z³ - 3xyz is?

The sum of $\frac{1}{x + y}$ $\frac{1}{{x + y}}$ and $\frac{1}{x - y}$ $\frac{1}{{x - y}}$ is?

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algebra - 05

01. The simplified value of following is: (315a5b6c3×59ab5c4)(315a5b6c3×59ab5c4)\left( {\frac{3}{{15}}{a^5}{b^6}{c^3} \times \frac{5}{9}a{b^5}{c^4}} \right) ÷ 1017a2bc31017a2bc3\frac{{10}}{{17}}{a^2}b{c^3}

02. If a2 + b2 + c2 = 2(a - b - c) -3, then the value of a + b + c is?

03. If x(x + y + z) = 20, y = (x + y + z) = 30 & z(x + y + z) = 50, then the value of 2(x + y + z) is?

04. If x = y = z, then (x+y+z)2x2+y2+z2(x+y+z)2x2+y2+z2\frac{{{{\left( {x + y + z} \right)}^2}}}{{{x^2} + {y^2} + {z^2}}} is?

05. If xy = a+2a−2,xy = a+2a−2,\frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}}{\text{,}} then the value of x2−y2x2+y2x2−y2x2+y2\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} = ?

06. If x−1x=2,x−1x=2,x - \frac{1}{x} = 2{\text{,}} then the value of the following is x3−1x3x3−1x3{x^3} - \frac{1}{{{x^3}}} = ?

07. If a2 + b2 + c2 - ab - bc - ca = 0 then a : b : c is?

08. If x+1x−1=abx+1x−1=ab\frac{{x + 1}}{{x - 1}} = \frac{a}{b} and 1−y1+y=ba,1−y1+y=ba,\frac{{1 - y}}{{1 + y}} = \frac{b}{a}{\text{,}} then the value of x−y1+xyx−y1+xy\frac{{x - y}}{{1 + xy}} is?

09. If x + y + z = 6 and xy + yz + zx = 10, then the value of x3 + y3 + z3 - 3xyz is?

10. The sum of 1x+y1x+y\frac{1}{{x + y}} and 1x−y1x−y\frac{1}{{x - y}} is?

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The simplified value of following is: $(\frac{3}{15} a^{5} b^{6} c^{3} \times \frac{5}{9} a b^{5} c^{4})$ $\left( {\frac{3}{{15}}{a^5}{b^6}{c^3} \times \frac{5}{9}a{b^5}{c^4}} \right)$ ÷ $\frac{10}{17} a^{2} b c^{3}$ $\frac{{10}}{{17}}{a^2}b{c^3}$

02. If a² + b² + c² = 2(a - b - c) -3, then the value of a + b + c is?

If x = y = z, then $\frac{{(x + y + z)}^{2}}{x^{2} + y^{2} + z^{2}}$ $\frac{{{{\left( {x + y + z} \right)}^2}}}{{{x^2} + {y^2} + {z^2}}}$ is?

If $\frac{x}{y} = \frac{a + 2}{a - 2},$ $\frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}}{\text{,}}$ then the value of $\frac{x^{2} - y^{2}}{x^{2} + y^{2}}$ $\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}$ = ?

If $x - \frac{1}{x} = 2,$ $x - \frac{1}{x} = 2{\text{,}}$ then the value of the following is $x^{3} - \frac{1}{x^{3}}$ ${x^3} - \frac{1}{{{x^3}}}$ = ?

07. If a² + b² + c² - ab - bc - ca = 0 then a : b : c is?

If $\frac{x + 1}{x - 1} = \frac{a}{b}$ $\frac{{x + 1}}{{x - 1}} = \frac{a}{b}$ and $\frac{1 - y}{1 + y} = \frac{b}{a},$ $\frac{{1 - y}}{{1 + y}} = \frac{b}{a}{\text{,}}$ then the value of $\frac{x - y}{1 + x y}$ $\frac{{x - y}}{{1 + xy}}$ is?

09. If x + y + z = 6 and xy + yz + zx = 10, then the value of x³ + y³ + z³ - 3xyz is?

The sum of $\frac{1}{x + y}$ $\frac{1}{{x + y}}$ and $\frac{1}{x - y}$ $\frac{1}{{x - y}}$ is?

NCERT
6 to 12