Question

If x² + 1/ x² = 38 then find the value of x³- 1/ x³​

Answer 1

${\large{\underbrace{\underline{\bf{Required\;Answer:-}}}}}$

» We Know :

• $\sf \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 38 \: \dots(i)$

» Finding $\bf x-\dfrac{1}{x}$ :

$\displaystyle \colon \rarr \: \: \sf \bigg (x - \frac{1}{x} \bigg ) {}^{ {}^{ \displaystyle2} } = {x}^{2} + \frac{1}{ {x}^{2} } - 2( \cancel{x}) \bigg( \frac{1}{ \cancel{x}} \bigg)$

• From equation (i) , $\tt x^2+\dfrac{1}{x^2}=38$

$\\ \colon \rarr \: \: \sf \bigg(x - \dfrac{1}{x} \bigg) {}^{ {}^{ \displaystyle2} } = 38 - 2$

$\colon \rarr \: \sf \: \bigg(x - \dfrac{1}{x} \bigg) {}^{ {}^{ \displaystyle2} } = 36$

$\colon \rarr \: \sf \: \bigg(x - \dfrac{1}{x} \bigg) = \sqrt{36}$

$\colon \rarr \: \sf \: \boxed{ \bigg( \bf x - \dfrac{1}{x} \bigg) = 6} \dots(ii)$

» Now :

Using the formula -

$\quad \quad \quad \star \boxed{ \tt {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + {b}^{2} + ab)}$

Put -

• a = x
• b = $\sf\dfrac{1}{x}$

$\colon \implies \sf \: {x}^{3} - \dfrac{1}{ {x}^{3} } = \underline{ \bigg(x - \dfrac{1}{x} \bigg) }\bigg \{ \underline{ {x}^{2} + \frac{1}{ {x}^{2} } } + ( \cancel{x}) \bigg( \frac{1}{ \cancel{x}} \bigg) \bigg \}$

• From equation (i) , $\tt x^2+\dfrac{1}{x^2}=38$

• From equation (ii) , $\tt x-\dfrac{1}{x}=6$

$\\ \colon \implies \: \sf\: {x}^{3} - \frac{1}{ {x}^{3} } = 6 \times (38 + 1)$

$\colon \implies \: \sf \: {x}^{3} + \frac{1}{ {x}^{3} } = 6 \times 39$

$\colon \dashrightarrow \underline{ \boxed{ \bf{ {x}^{3} - \frac{1}{ {x}^{3} } = 234}}}$

So,

» Final Answer :

★ The value of $\bf x^3-\dfrac{1}{x^3}$ would be 234 .

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