Question

if x+1/x = 3 then what is x²+1/x²​

Answer 1

Given :-

$\sf\bigg(x+\dfrac{1}{x}\bigg)$

To Find :-

$\bigg(x^2+\dfrac{1}{x^2}\bigg)$

Solution :-

We know that

(a + b)² = a² + 2ab + b²

$\sf\bigg(x+\dfrac{1}{x}\bigg)$

$\sf \bigg(x+\dfrac{1}{x^2}\bigg)$

$\sf x^2+2\times x^2\times\dfrac{1}{x^2}+\dfrac{1}{x^2}=(3)^2$

$\sf x^2+\dfrac{1}{x^2}+2=9$

$\sf x^2+\dfrac{1}{x^2}=9-2$

$\sf x^2+\dfrac{1}{x^2}=7$

Answer 2

$\large \bigstar \: \sf \underline\bold{ \underline{Given : }}$

$\sf \bold{\bigg(x + \frac{1}{x} = 3\bigg)}$

$\large\bigstar \: \sf \underline\bold{ \underline{To \: find: }}$

$\sf \bold{\bigg({x}^{3} + \frac{1}{ {x}^{3} } \bigg)}$

$\large \bigstar \: \sf \underline\bold{ \underline{Solution: }}$

$: \implies \: \red{\boxed{ {x}^{3} + \frac{1}{ {x}^{3} } = 18} \: } \green\bigstar$

Formula used:-

$</p><p>\underline{\boxed{\sf{\pm{ \bold{\green{{(a + b)}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b) }}}}}}\:\red{\bigstar}$

Given that,

$: \implies \sf \bold \: x + \frac{1}{x} = 3$

$\sf \bold\blue {\underline{Taking \: cube \: on \: both \: sides,}}$

$: \implies \sf \bold \: { \bigg(x + \frac{1}{x} \bigg)}^{3} = {(3)}^{3}$

$\sf \bold \orange{ \underline{Using \: the \: given \: formula,}}$

$: \implies \sf \bold \: {\small {x}^{3} + { \big( \frac{1}{x} \big)}^{3} + 3 \times x \times \frac{1}{x} (x + \frac{1}{x} ) = 27}$

$: \implies \sf \bold \: {\rightarrow \: {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times (3) = 27 }$

$: \implies \sf \bold \: {\because \: x + \frac{1}{x} = 3 }$

$:\implies \:\underline{\boxed{\frak{\pm{\pink{{x}^{3} + \frac{1}{ {x}^{3} } = 18}}}}}\:\red{\bigstar}$