Question

if x-1/x=5 ,the value of x^3 -1/x^3 is​

Answer 1

$\bf Given$

$\tt\to x-\dfrac{1}{x}=5$

$\bf To\:\:Find$

$\tt\to x^{3} -\dfrac{1}{x^3}$

$\bf Using\:\:\:This\:\:\: Formula$

$\tt\to( a - b)^3= a^3-b^3-3ab(a-b)$

$\bf Cube \:\:on\:\: both\:Side$

$\tt\to \bigg(x-\dfrac{1}{x}\bigg)^3=5^3$

$\tt\to x^3-\dfrac{1}{x^3}-3\times x\times \dfrac{1}{x} \bigg( x - \dfrac{1}{x}\bigg)=125$

$\tt\to x^3-\dfrac{1}{x^3} -3(5) = 125$

$\tt\to x^3-\dfrac{1}{x^3} -15 = 125$

$\tt\to x^3-\dfrac{1}{x^3} = 125 +15$

$\tt\to x^3-\dfrac{1}{x^3} = 140$

$\bf Answer$

$\tt\to x^3-\dfrac{1}{x^3} = 140$

Answer 2

Solution

Given :-

• x - 1/x = 5 ____________(1)

Find :-

• x³ - 1/x³ = ?

Explanation

Using Formula

★ (a + b)² = a² + b² + 2ab

★(a³ - b³) = (a - b)(a² + ab + b²)

Now, squaring both side of equ(1)

==> (x - 1/x)² = 5²

==> x² + 1/x² - 2 × x × 1/x = 25

==> x² + 1/x² - 2 = 25

==> x² + 1/x² = 25 + 2

==> x² + 1/x² = 27________________(2)

Now, calculate

==> x³ - 1/x³ = ( x - 1/x)(x² + 1/x² + 1)

==> keep Value by equ (1) & (2)

==> x³ - 1/x³ = 5 × (27 + 1)

==> x³ - 1/x³ = 5 × 28

==> x³ - 1/x³ = 140

Hence

• Value will be x³ - 1/x³ = 140

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