x 3 - 1 x 3 x 3 - 1 x 3 - xk-1xk where k is -

(a+b)(a b)=a^2 ab+ab b^2=a^2 b^2 ∴ #10;[x^3 (1/x^3)] · [x^3+(1/x^3)]=[x^6 (1/x^6)] ∴ k=6 ...

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Byju's Answer

Standard VIII

Mathematics

Addition and Subtraction of Algebraic Expression

x 3 + 1 x 3...

Question

$(x^{3} + \frac{1}{x^{3}}) (x^{3} - \frac{1}{x^{3}}) =$ $(x^{k} - \frac{1}{x^{k}})$ where k is $?$

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Solution

$(a + b) (a - b) = a^{2} - a b + a b - b^{2} = a^{2} - b^{2} ∴ [x^{3} - (\frac{1}{x^{3}})] \cdot [x^{3} + (\frac{1}{x^{3}})] = [x^{6} - (\frac{1}{x^{6}})]$
$∴$ $k = 6$

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(x3+1x3)(x3−1x3)= (xk−1xk) where k is?

(a+b)(a−b)=a2−ab+ab−b2=a2−b2∴[x3−(1x3)]⋅[x3+(1x3)]=[x6−(1x6)]∴ k=6

$(x^{3} + \frac{1}{x^{3}}) (x^{3} - \frac{1}{x^{3}}) =$ $(x^{k} - \frac{1}{x^{k}})$ where k is $?$

$(a + b) (a - b) = a^{2} - a b + a b - b^{2} = a^{2} - b^{2} ∴ [x^{3} - (\frac{1}{x^{3}})] \cdot [x^{3} + (\frac{1}{x^{3}})] = [x^{6} - (\frac{1}{x^{6}})]$
$∴$ $k = 6$