Question

$if \: \frac{x { }^{2} - 1}{x} = 2 \: then \: find \: out \: \frac{ {x}^{6} - 1 }{x3} = how \: mch$
​

Answer 1

Given :-

• (x² - 1)/x = 2 .

To Find :-

• (x^6 - 1) / x³ = ?

Solution :-

→ (x² - 1)/x = 2

Taking x common from LHS Numerator , we get,

→ x[x - (1/x) ] / x = 2

→ (x - 1/x) = 2 ---------- Equation (1)..

____________

Now, we have to Find :-

→ (x^6 - 1) / x³

Taking x³ common from LHS Numerator , we get,

→ x³( x³ - 1/x³) / x³

→ (x³ - 1/x³) = ? ----------- Equation (2).

_____________

Cubing Both Sides of Equation (1) now, we get,

→ (x - 1/x)³ = 2³

using (a - b)³ = a³ - b³ - 3ab(a - b) we get,

→ x³ - 1/x³ - 3 * x * 1/x (x - 1/x) = 8

→ x³ - 1/x³ - 3(x - 1/x) = 8

Putting value of Equation (1) , again,

→ (x³ - 1/x³) - 3 * 2 = 8

→ (x³ - 1/x³) = 8 + 6

→ (x³ - 1/x³) = 14 = Equation (2). = Our Answer.

Answer 2

Answer:

$\underline{\bigstar\:\textbf{We Have Given :}}$

$:\implies\sf \dfrac{x^2 - 1}{x} = 2\\\\{\scriptsize\qquad\bf{\dag}\:\:\texttt{Taking x Common}}\\\\:\implies\sf x\Bigg\lgroup\dfrac{x-\frac{1}{x}}{x}\Bigg\rgroup=2\\\\\\:\implies\sf x - \dfrac{1}{x} = 2$

$\rule{130}{1}$

$\underline{\bigstar\:\textbf{According to the Question :}}$

$:\implies\sf \dfrac{x^6- 1}{x^3}\\\\{\scriptsize\qquad\bf{\dag}\:\:\tt{Taking\:x^3\:Common}}\\\\:\implies\sf x^3\Bigg\lgroup\dfrac{x^3-\frac{1}{x^3}}{x^3}\Bigg\rgroup \\\\\\:\implies\sf x^3 - \dfrac{1}{x^3}\\\\{\scriptsize\qquad\bf{\dag}\:\:\tt{x-\frac{1}{x}=a\quad then \quad x^2-\frac{1}{x^3}=(a^3+3a)}}\\\\:\implies\sf x^3-\dfrac{1}{x^3}=(2)^3+3(2)\\\\\\:\implies\sf x^3-\dfrac{1}{x^3} = 8 + 6\\\\\\:\implies\underline{\boxed{\sf x^3-\dfrac{1}{x^3} = 14}}$

⠀

$\therefore\:\underline{\textsf{Hence, Required value will be \textbf{14}}}.$