Question

only for bainly people ......................

Answer 1

Answer: 2350

Step-by-step explanation:  x + 1/x = 5 ( given )

Find the value of x⁵ + 1/x⁵ = ?

So,

x + 1/x = 5 ...(1)

Squaring both sides

( x+1/x)² = 5²

x² + 1/x² + 2  =25

x² + 1/x² = 25 - 2

x² + 1/x ² = 23  ....(2)

Cube of the equation

x³ + 1/x³ = ( x+ 1/x )³ - 3 (x +1/x)

x³ + 1/x ³ = (5)³ - 3 (5)

x³ + 1/x³ = 125 -15

x³ + 1/x³ = 110 ...(3)

Now taking all equation

(x² + 1/x² ) (x³ + 1/x³ ) = (x⁵ +1/x +x + 1/x⁵)

23 ₓ 110 = x⁵ + 1/x⁵

2530 = x⁵ + 1/x⁵

Answer 2

Answer:

7950

Step-by-step explanation:

Given

$x + \frac{1}{x}=6$

$x^{2} + \frac{1}{x^{2}}=?$

Now given is

$x + \frac{1}{x}=6$         ..............(i)

Taking square of both sides

$(x + \frac{1}{x})^{2}=6^{2}$

we know that $(a+b)^{2}=a^{2}+b^{2}+2ab$

so the given becomes

$x^{2} + \frac{1}{x^{2}}+2=36$

$x^{2} + \frac{1}{x^{2}}=36-2$

$x^{2} + \frac{1}{x^{2}}=34$                ...............(ii)

Now Taking cube of the equation (i)

$(x + \frac{1}{x})^{3}=6^{3}$

the formula for cube is $(a+b)^{3}=a^{3}+b^{3}-3(a+b)($

The equation becomes

$x^{3}+\frac{1}{x^{3} }-3(x+\frac{1}{x})=216$

Also the value of  x+\frac{1}{x} is given so putting it

$x^{3}+\frac{1}{x^{3} }-3(6)=216$

$x^{3}+\frac{1}{x^{3} }-18)=216$

$x^{3}+\frac{1}{x^{3} })=216+18$

$x^{3}+\frac{1}{x^{3} })=234$                ...............(iii)

Now to get value of it

multiplying eq (ii) with eq(iii)

$(x^{2}+\frac{1}{x^{2} }) * (x^{3}+\frac{1}{x^{3} })= 234*34$

$(x^{2}+\frac{1}{x^{2} }) * (x^{3}+\frac{1}{x^{3} })= 7956$

$x^{5}+\frac{1}{x^{5} } + x +\frac{1}{x}=7956$

putting the known values

$x^{5}+\frac{1}{x^{5} } +6=7956$

$x^{5}+\frac{1}{x^{5} }=7956-6$

$x^{5}+\frac{1}{x^{5} }=7950$