Question

If x^4+1/x^4=47 find the value of x^3+1/x^3​

Answer 1

Answer:

18

Step-by-step explanation:

Given :

To find the value of :

$x^3+\frac{1}{x^3}$ if,

$x^4+\frac{1}{x^4}=47$

Solution :

$x^4+\frac{1}{x^4}=47$

By adding 2 both the sides,

⇒ $x^4+\frac{1}{x^4} + 2=47 + 2$

⇒ $x^4+\frac{1}{x^4} + 2(\frac{x^2}{x^2})=49$

⇒ $(x^2)^2+(\frac{1}{x^2})^2 + 2(x^2)(\frac{1}{x^2})=7 \times 7$

It is of the form,

(a + b)² = a² + 2ab + b²

Here, $a=x^2$ , $b=\frac{1}{x^2}$

So, We get,

⇒ $(x^2 + \frac{1}{x^2})^2 =7^2$

By taking root both the sides,

We get,

$x^2 + \frac{1}{x^2} =7$ ...(i)

By adding 2 both the sides,

⇒ $x^2+\frac{1}{x^2} + 2=7 + 2$

⇒ $x^2+\frac{1}{x^2} + 2(\frac{x}{x})=9$

⇒ $(x)^2+(\frac{1}{x})^2 + 2(x)(\frac{1}{x})=3 \times 3$

It is of the form,

(a + b)² = a² + 2ab + b²

Here, $a=x$ , $b=\frac{1}{x}$

So, We get,

⇒ $(x + \frac{1}{x})^2 =3^2$

By taking root both the sides,

We get,

$x + \frac{1}{x} =3$ ...(ii)

We know that,

$a^3+b^3 = (a+b)(a^2-ab+b^2)$

By substituting,

$a=x$ , $b=\frac{1}{x}$,

We get,

⇒ $x^3+\frac{1}{x^3}=(x+\frac{1}{x})[x^2 - (x)(\frac{1}{x})+\frac{1}{x^2}]$

⇒ $x^3+\frac{1}{x^3}=(x+\frac{1}{x})[x^2 - 1 +\frac{1}{x^2}]$

⇒ $x^3+\frac{1}{x^3}=(x+\frac{1}{x})[x^2 +\frac{1}{x^2} - 1]$

⇒ $x^3+\frac{1}{x^3}=(3)[7 - 1] = 3 \times 6 = 18$