Question

2
If x - 6x + 1 = 0 , then the value of
${x}^{3} + \frac{1}{ {x}^{3} }$
can be​

Answer 1

$\boxed{ \red{AnswEr:-}}$

• 198

★ Correct Question :-

• If x² - 6x + 1 = 0, then what is the value of x³ + (1/x)³

★ GivEn :-

• $\sf \: {x}^{2} - 6x + 1 = 0$

★ To Find :-

• $\sf \: value \: of \: {x}^{3} + \dfrac{1}{ {x}^{3} }$

$\boxed{ \red{ Solution :-}}$

$\implies \sf \: {x}^{2} - 6x + 1 = 0$

$\implies \sf \: {x}^{2} - 6x = - 1$

$\implies \sf \: x(x - 6) = - 1$

$\implies \sf \: x - 6 = - \dfrac{ 1}{x}$

$\implies \boxed{\sf \: {\bold{ \red{ x + \dfrac{1}{x} = 6}}}}$

Now, ➦

$\sf \implies {x}^{3} + \dfrac{1}{ {x}^{3}}$

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

$\implies \sf \: {6}^{3} - 3 \times 6$

$\implies \sf \: 216 - 18$

$\implies \boxed{ \sf{ \red{ \bold{198}}}}$

Answer 2

Correct Question :

If x² - 6x + 1 = 0, then what is the value of $\sf x^{3} + \dfrac{1}{x^{3}}$

Given :

$\sf {x}^{2} - 6x + 1 = 0$

To Find :

$\sf The \: value \: of \: {x}^{3} + \dfrac{1}{ {x}^{3} }$

Solution :

As, we are given :

$\sf {x}^{2} - 6x + 1 = 0$

$\sf : \implies {x}^{2} - 6x = - 1$

$\sf : \implies x(x - 6) = - 1$

$\sf : \implies x - 6 = - \dfrac{ 1}{x}$

$\sf : \implies x + \dfrac{1}{x} = 6$

$\underline{\boxed{\bf x + \dfrac{1}{x} = 6}}$

By cubing both sides, we get :

$\sf : \implies \Bigg(x + \dfrac{1}{x} \Bigg)^{3} = (6)^{3}$

By using identinty :

$\underline{\boxed{\bf{ (a+b)^{3} = a^{3} + b^{3} + 3a^{2}b + 3ab^{2} }}}$

$\sf : \implies x^{3} + \Bigg(\dfrac{1}{x}\Bigg)^{3} + 3 \times x^{2} \times \dfrac{1}{x} + 3 \times x \times (\dfrac{1}{x} )^{2} = 216$

$\sf : \implies x^{3} + \dfrac{1}{x^{3}} + 3x + \dfrac{3}{x} = 216$

$\sf : \implies x^{3} + \dfrac{1}{x^{3}} + 3 \Bigg(x + \dfrac{1}{x}\Bigg) = 216$

Now, we have :

$\underline{\boxed{\bf x + \dfrac{1}{x} = 6}}$

$\sf : \implies x^{3} + \dfrac{1}{x^{3}} + 3 \times 6 = 216$

$\sf : \implies x^{3} + \dfrac{1}{x^{3}} + 18 = 216$

$\sf : \implies x^{3} + \dfrac{1}{x^{3}} + 18 = 216$

$\sf : \implies x^{3} + \dfrac{1}{x^{3}} = 216 - 18$

$\sf : \implies x^{3} + \dfrac{1}{x^{3}} = 198$

$\Large \underline{\boxed{\bf{x^{3} + \dfrac{1}{x^{3}} = 198}}}$

$\pink{ \sf \therefore \: value \: of \: x^{3} + \dfrac{1}{x^{3}} = 198}$