Product of x-1x x-1x x2-1x2 is -

The correct option is C x^4 1x^4 (a+b)(a b)=a^2 b^2 By this property, ( x 1 x #160; ) ( x+ 1 x #160; ) =( x^ 2 1 x^ 2 #160; ) ...

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Byju's Answer

Standard XII

Mathematics

Sum of Coefficients of All Terms

Product of ...

Question

Product of $(x - \frac{1}{x}) (x + \frac{1}{x}) (x^{2} + \frac{1}{x^{2}})$ is :

x^{4} + \frac{1}{x^{4}}

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

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x^{4} - \frac{1}{x^{4}}

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

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Solution

The correct option is C $x^{4} - \frac{1}{x^{4}}$
$(a + b) (a - b) = a^{2} - b^{2}$
By this property, $(x - \frac{1}{x}) (x + \frac{1}{x}) = (x^{2} - \frac{1}{x^{2}}) \Rightarrow (x - \frac{1}{x}) (x + \frac{1}{x}) (x^{2} + \frac{1}{x^{2}}) = (x^{2} - \frac{1}{x^{2}}) (x^{2} + \frac{1}{x^{2}}) = (x^{4} - \frac{1}{x^{4}})$

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Similar questions

If $x + \frac{1}{x} = 3$ , calculate $x^{2} + \frac{1}{x^{2}}, x^{3} + \frac{1}{x^{3}} a n d x^{4} + \frac{1}{x^{4}}$ .

State true or false:

x \neq 0

and

x + \frac{1}{x} = 2

, then:

x^{2} + \frac{1}{x^{2}} = x^{3} + \frac{1}{x^{3}} = x^{4} + \frac{1}{x^{4}}

If $x_{1}, x_{2}, x_{3}, x_{4}$ are four positive real numbers such that $x_{1} + \frac{1}{x_{2}} = 4, x_{2} + \frac{1}{x_{3}} = 1, x_{3} + \frac{1}{x_{4}} = 4$ and $x_{4} + \frac{1}{x_{1}} = 1,$ then

Q. Determine which of the following polynomials has

(x + 1)

a factor:
(i)

x^{3} + x^{2} + x + 1

(ii)

x^{4} + x^{3} + x^{2} + x + 1

(iii)

x^{4} + 3 x^{3} + 3 x^{2} + x + 1

(iv)

x^{3} - x^{2} - (2 + \sqrt{2}) x + \sqrt{2}

Q. Which of the following polynomials has (x + 1) as a factor?
I.

x^{3} + x^{2} + x + 1

II.

x^{4} + x^{3} + x^{2} + x + 1

III.

x^{4} + 3 x^{3} + 3 x^{- 2} + x + 1

II.

x^{3} + x^{2} - (2 + \sqrt{2}) x + \sqrt{2}

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Product of (x−1x)(x+1x)(x2+1x2) is :

The correct option is C x4−1x4(a+b)(a−b)=a2−b2By this property, (x−1x)(x+1x)=(x2−1x2)⇒(x−1x)(x+1x)(x2+1x2)=(x2−1x2)(x2+1x2)=(x4−1x4)

Product of $(x - \frac{1}{x}) (x + \frac{1}{x}) (x^{2} + \frac{1}{x^{2}})$ is :