Question

Please solve the problem in above of question​

Answer 1

Given Equation

$\tt \to \: \bigg( {x}^{3} - \dfrac{1}{ {x}^{3} } \bigg) \bigg( {x}^{3} + \dfrac{1}{ {x}^{3} } \bigg)\bigg( {x}^{6} + \dfrac{1}{ {x}^{6} } \bigg)$

To Prove

$\to \tt\bigg( {x}^{12} - \dfrac{1}{ {x}^{12} } \bigg)$

Now Take

$\tt \to \: \bigg( {x}^{3} - \dfrac{1}{ {x}^{3} } \bigg) \bigg( {x}^{3} + \dfrac{1}{ {x}^{3} } \bigg)\bigg( {x}^{6} + \dfrac{1}{ {x}^{6} } \bigg)$

By using this identity

$\tt \to \:(a + b)(a - b) = {a}^{2} - {b}^{2}$

we get

$\tt \to \: \bigg( \{{x}^{3} \} {}^{2} - \dfrac{1}{ \{{x}^{3} \}^{2} } \bigg) \bigg( {x}^{6} + \dfrac{1}{ {x}^{6} } \bigg)$

$\tt \to \: \bigg( {x}^{3 \times 2} - \dfrac{1}{ {x}^{3 \times 2} } \bigg) \bigg( {x}^{6} + \dfrac{1}{ {x}^{6} } \bigg)$

$\tt \to \: \bigg( {x}^{6} - \dfrac{1}{ {x}^{6} } \bigg) \bigg( {x}^{6} + \dfrac{1}{ {x}^{6} } \bigg)$

Again using same identity

$\tt \to \:(a + b)(a - b) = {a}^{2} - {b}^{2}$

we get

$\tt \to \: \bigg( \{{x}^{6} \} {}^{2} - \dfrac{1}{ \{{x}^{6} \}^{2} } \bigg)$

$\tt \to \: \bigg( {x}^{6\times 2} - \dfrac{1}{ {x}^{6 \times 2} } \bigg)$

$\tt \to \: \bigg( {x}^{ 12} - \dfrac{1}{ {x}^{12} } \bigg)$

LHS = RHS

Hence Proved

Answer 2

Answer:

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