Question

if x is equals to 7 + 4 root 3 find the value of bracket x cube + 1 by x cube bracket

Answer 1

Here's your answer...

Answer 2

Answer:

The required value is $(x^3+\frac{1}{x^3})=2702$

Step-by-step explanation:

Given : $x=7+4\sqrt3$

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

Solution :

We know, $x=7+4\sqrt3$ .....(1)

$\frac{1}{x}=\frac{1}{7+4\sqrt3}$

Rationalizing,

$\frac{1}{x}=\frac{1}{7+4\sqrt3}\times \frac{7-4\sqrt3}{7-4\sqrt3}$

$\frac{1}{x}=\frac{7-4\sqrt3}{7^2-(4\sqrt3)^2}$

$\frac{1}{x}=\frac{7-4\sqrt3}{49-48}$

$\frac{1}{x}=7-4\sqrt3$ ....(2)

Cubing equation (1),

$x^3=(7+4\sqrt3)^3$

$x^3=(7)^3+(4\sqrt3)^3+3(7)^2(4\sqrt3)+3(7)(4\sqrt3)^2$

$x^3=343+192\sqrt3+588\sqrt3+1008$

$x^3=1351+780\sqrt3$ ......(3)

Cubing equation (2),

$\frac{1}{x^3}=(7-4\sqrt3)^3$

$\frac{1}{x^3}=(7)^3+(-4\sqrt3)^3+3(7)^2(-4\sqrt3)+3(7)(-4\sqrt3)^2$

$\frac{1}{x^3}=343-192\sqrt3-588\sqrt3+1008$

$\frac{1}{x^3}=1351-780\sqrt3$ ......(4)

Adding (3) and (4),

$x^3+\frac{1}{x^3}=1351+780\sqrt3+1351-780\sqrt3$

$x^3+\frac{1}{x^3}=2702$

Therefore, The required value is $(x^3+\frac{1}{x^3})=2702$