Question

x⁴+1uponx⁴= 119 find the value of x³-1/x³​

Answer 1

★ᏀᏆᐯᗴᑎ:-

$\sf x^4 + \dfrac{1}{x^4} = 119$

★Ͳϴ ҒᏆΝᎠ:-

$\sf x - \dfrac{1}{x}$

★ᔑᝪしᑌᎢᏆᝪᑎ:-

$\sf (a+b)^2 = a^2 + b^2 + 2ab$

First, we will find $x^2 + \dfrac{1}{x^2}$

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

$\sf \implies \Bigg ( x^2 + \dfrac{1}{x^2} \Bigg )^2= x^4+ \dfrac{1}{x^4} + 2$

$\sf \implies \Bigg ( x^2 + \dfrac{1}{x^2} \Bigg )^2= 119 + 2$

$\sf \implies \Bigg ( x^2 + \dfrac{1}{x^2} \Bigg )^2= 121$

$\sf \implies x^2 + \dfrac{1}{x^2}= \sqrt{121}$

$\sf \implies x^2 + \dfrac{1}{x^2}= 11$

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Now, we will find $x- \dfrac{1}{x}$

$\sf (a-b)^2 = a^2+b^2-2ab$

$\sf \implies \Bigg ( x - \dfrac{1}{x} \Bigg )^2 = (x)^2 + \Bigg ( \dfrac{1}{x} \Bigg )^2 - 2 \times x \times \dfrac{1}{x}$

$\sf \implies \Bigg ( x - \dfrac{1}{x} \Bigg )^2 = x^2 + \dfrac{1}{x^2} - 2$

$\sf \implies \Bigg ( x - \dfrac{1}{x} \Bigg )^2 = 11 - 2$

$\sf \implies \Bigg ( x - \dfrac{1}{x} \Bigg )^2 = 9$

$\sf \implies x - \dfrac{1}{x} = \sqrt{9}$

$\sf \implies x - \dfrac{1}{x} = 3$

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Finally, we will find $\sf x^3 - \dfrac{1}{x^3}$

We know that a³-b³ = (a-b)(a²+b²+ab)

Using this identity,

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

$\sf \implies x^3 - \dfrac{1}{x^3} = (3) ( 11+ 1)$

$\sf \implies x^3 - \dfrac{1}{x^3} = (3) ( 12)$

$\sf \implies x^3 - \dfrac{1}{x^3} = 36$

Hope it helps.

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