Question

If x²+1/x²=83,find value of x³-1/x³

Answer 1

$= {x}^{2} + 1 \div {x}^{2} = 83 \\ = {x}^{2} + 1 \div {x}^{2} - 2 + 2 ( = 0) = 83 \\ = ( {x - 1 \div x})^{2} = 83 - 2 = 81 \\ = x - 1 \div x = 9 \\ = taking \: cube \: in \: both \: sides \\ = {x}^{3} - 1 \div {x}^{3} - 3(x - 1 \div x) = 729 \\ = {x}^{3} - 1 \div {x}^{3} - 3(9) = 729 \\ = {x}^{3} - 1 \div {x}^{3} = 729 - 27 \\ =( {x}^{3} - 1 \div {x}^{3} ) = 702 \: ans$

Answer 2

Answer :

We know that

$\\ \qquad \sf \bigg(x - \frac{1}{x} \bigg){}^{2} = x {}^{2} + \frac{1}{ {x}^{2} } - 2 \\ \\ \\ \implies \sf \bigg(x - \frac{1}{x} \bigg) {}^{2} = 83 - 2 \qquad \bigg \{putting \: {x}^{2} + \frac{1}{ {x}^{2} } = 83\bigg \} \\ \\ \\ \implies \sf \bigg(x - \frac{1}{x} \bigg) {}^{2} = 81 \\ \\ \\ \implies \sf \bigg(x - \frac{1}{x} \bigg) {}^{2} = {9}^{2} \\ \\ \\ \implies \sf x - \frac{1}{x} = 9 \\ \\ \\ \implies \sf \bigg(x - \frac{1}{x} \bigg) {}^{3} = {9}^{3} \\ \\ \\ \implies \sf {x}^{3} - \frac{1}{ {x}^{3} } - 3 \: \bigg(x - \frac{1}{x} \bigg) = 729 \\ \\ \\ \implies \sf {x}^{3} - \dfrac{1}{ {x}^{3} } - 3 \times 9 = 729 \\ \\ \\ \implies \sf {x}^{3} - \frac{1}{ {x}^{3} } = 729 + 27 \\ \\ \\ \implies \sf \blue{ {x}^{3} - \frac{1}{ {x}^{3} } = 756}$