Find the cube of x-1-xA- x3-1-x3-3 x-3-x3B- x3-1-x3-3 x-3-x2C- x3-1-x3-3 x2-3-xD- x3-1-x3-3 x2-3-x2

The correct option is A x^3+1/x^3+3x+3/x Cube of x+1/x=(x+1/x)^3 Using identity: (a+b)^3=a^3+b^3+3a^2b+3ab^2, we get (x+1/x)^3=x^3+1/x^3+3× x^2 ×1/x+3× x ×1/x^ ...

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Byju's Answer

Standard VIII

Mathematics

Algebraic Identity (a+b)^3

Find the cube...

Question

Find the cube of $x + \frac{1}{x}$

$x^{3} + \frac{1}{x^{3}} + 3 x^{2} + \frac{3}{x^{2}}$

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$x^{3} + \frac{1}{x^{3}} + 3 x + \frac{3}{x}$

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$x^{3} + \frac{1}{x^{3}} + 3 x + \frac{3}{x^{2}}$

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$x^{3} + \frac{1}{x^{3}} + 3 x^{2} + \frac{3}{x}$

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Solution

The correct option is B
$x^{3} + \frac{1}{x^{3}} + 3 x + \frac{3}{x}$

Cube of $x + \frac{1}{x} = {(x + \frac{1}{x})}^{3}$

Using identity: $(a + b)^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}$ , we get

${(x + \frac{1}{x})}^{3} = x^{3} + \frac{1}{x^{3}} + 3 \times x^{2} \times \frac{1}{x} + 3 \times x \times \frac{1}{x^{2}}$

$= x^{3} + \frac{1}{x^{3}} + 3 x + \frac{3}{x}$

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Similar questions

Q. The polynomial which when divided by −x² + x − 1 gives a quotient x − 2 and remainder 3, is

(a) x³ − 3x² + 3x − 5
(b) −x³ − 3x² − 3x − 5
(c) −x³ + 3x² − 3x + 5
(d) x³ − 3x² − 3x + 5

Q. Evaluate :

\int \frac{d x}{(x^{3} + 3 x^{2} + 3 x + 1) \sqrt{x^{2} + 2 x - 3}}

Q. Solve:

y = {tan}^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}), - \frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}

Q. Match the following.

	Column - I		Column - II
(P)	$(x^{2} + 5) (x^{3} + 3) + 5$	(1)	${- x}^{3} - 3 x^{2} + 3 x + 2$
(Q)	$(\frac{- 10}{3} x y^{3}) \times (\frac{6}{5} x^{3} y)$	(2)	$- x^{3} + x^{2} + 3 x - 6$
(R)	$(x^{3} - x^{2} - x - 2) - (2 x^{3} + 2 x^{2} - 4 x - 4)$	(3)	$x^{5} + 5 x^{3} + 3 x^{2} + 20$
(S)	$(x^{3} - x^{2} - x - 2) + (2 x^{2} - 2 x^{3} + 4 x - 4)$	(4)	$- 4 x^{4} y^{4}$

\int (x + x^{2} + x^{3} + x^{4})

dx is equal to

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Find the cube of x+1x

The correct option is B x3+1x3+3x+3x Cube of x+1x=(x+1x)3 Using identity: (a+b)3=a3+b3+3a2b+3ab2, we get (x+1x)3=x3+1x3+3×x2×1x+3×x×1x2 =x3+1x3+3x+3x

Find the cube of $x + \frac{1}{x}$