Question

find the value of x^3-1/x^3 if x+1/x=√29
(using the appropriate identities)​

Answer 1

Given,

$\boxed{ \huge\bf{x + \frac{1}{x} = \sqrt{29} }}$

We have,

• To Figure Out the value of x^3-(1/x^3)

Here,

• We need to know the value of x^2 + (1/x^2) {first}
• Then, Value of x - (1/x) has to be founded
• At last, using x - 1/x, we can find the value of x^3 - (1/x^3)

So,

$\to \: \rm{}x + \frac{1}{x} = \sqrt{29} \\ \\ \boxed{ \sf{}squaring \: \: on \: \: both \: \: sides} \\ \\ \to \rm { \bigg(x + \frac{1}{x} \bigg )}^{2} = {( \sqrt{29} )}^{2} \\ \\ \to \rm {x}^{2} + \frac{1}{ {x}^{2} } + 2 \bigg (\cancel{x}\bigg ) \bigg ( \frac{1}{ \cancel{x}} \bigg ) = 29\\ \\ \to \rm {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 29 \\ \\ \rm \to {x}^{2} + \frac{1}{ {x}^{2} } \: \: = 29 - 2 \: \: = \green{27}$

Then,

$\to \: \rm{}x - \frac{1}{x} \: \: \: \: \to {\bigg( {x}^{} - \frac{1}{x} \bigg)}^{2} \\ \\ \to \rm {(x)}^{2} + {\bigg( \frac{1}{x}\bigg )}^{2} - 2\bigg(x\bigg)\bigg( \frac{1}{x} \bigg) \\ \\ \rm \to \: {x}^{2} + \frac{1}{ {x}^{2} } - 2 \\ \\ \to \sf \: \: 27 - 2 \: \: \: \: \: \: \: \: \: \: \: \: \tiny \{ \bf using \: \: the \: \: obtained \: \: value \} \\ \to \sf \: \: {25} \\ \\ \therefore \: \: \: \rm x - \frac{1}{x} = \sqrt{25} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tiny \{ \bf \because 25 = {\big(x + \frac{1}{x} \big)}^{2} \} \\ \\ \implies \rm x - \frac{1}{x} = \pink{ \sf{5}}$

Now,

$\large \implies \rm \bigg(x - \frac{1}{x} \bigg ) = 5 \\ \\ \boxed {\tt{cubing \: \: on \: \: both \: \: sides} }\\ \\ \to \rm {(x)}^{3} - \frac{1}{ {( x )}^{3} } - 3 \bigg(\cancel{x} \bigg)\bigg( \frac{1}{\cancel{x}} \bigg)\bigg[x - \frac{1}{ {x}} \bigg] = {( \sqrt{29} )}^{3} \\ \\ \to \sf {x}^{3} - \frac{1}{ {x}^{3} } - 3\big[5\big] = 29 \sqrt{29} \\ \\ \to \sf{x}^{3} - \frac{1}{ {x}^{3} } - 15 = 29 \sqrt{29} \\ \\ \to \: \boxed{ \: \bf {x}^{3} - \frac{1}{ {x}^{3} } \: \: = \red{ 29 \sqrt{29} + 15} \: }$

Thus,

• We have obtained the required value of $\sf {x}^{3} - \frac{1}{ {x}^{3} }$ as $\red{\sf{29 \sqrt{29} + 15}}$.