Question

Pls answer me mates very very urgent

Answer 1

Given, x + (1/x) = 4.

On cubing both sides, we get

⇒ (x + 1/x)^3 = (4)^3

⇒ x^3 + 1/x^3 + 3(x + 1/x) = 64

⇒ x^3 + 1/x^3 + 3(4) = 64

⇒ x^3 + 1/x^3 + 12 = 64

⇒ x^3 + 1/x^3 = 64 - 12

⇒ x^3 + 1/x^3 = 52.

Therefore, the answer is Option (2) - cdba.

Hope it helps!

Answer 2

$<b><u><i>Hey mate!! Here's your answer</i></u></b>$
______________________________

$\huge{Given}$

$x + \frac{1}{x} = 4$

$\huge{To\: Find}$

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

$\huge{Solution}$

$x + \frac{1}{x} = 4$
Now, cubing both sides,

$( {x} + \frac{1}{x} )^{3} = ({4})^{3}$

$= > (x + \frac{1}{x} ) ^{3} = 64$


Now, using the formula (a+b)³ = a³+b³+3ab(a+b), we get


$= > x ^{ 3} + ( \frac{1}{x} ) ^{3} + 3 \times x \times \frac{1}{x} (x + \frac{1}{x} ) = 64$


Cancelling 3 × x × 1/x = 3 × 1 = 3


$= > {x}^{3} + ( \frac{1}{x} ) ^{3} + 3(x + \frac{1}{x} ) = 64$


Now, x + 1/x = 4. Then,


$= > x ^{3} + ( \frac{1}{x} ) ^{3} + 3(4) = 64$

$= > {x}^{3} + ( \frac{1}{x} ) ^{3} + 12 = 64$

$= > {x}^{3} + ( \frac{1}{x} ) ^{3} = 64 - 12$

$= > {x}^{3} + ( \frac{1}{x} ) ^{3} = 52$

Then, x³ + 1/x³ = 52

$\mathfrak{\blue{Finally,}}$

Your answer is -

$\boxed{\mathcal{\red{\bold{(2) \: cdba}}}}$
______________________________

$\huge{\mathfrak{\green{\bold{Thank \: you}}}}$