Question

If x4 + 1/x4 = 47, then find the value of x3 + 1/x3 op 1: 18 op 2: 27 op 3: 9 op 4: 12 op 5: correct op

Answer 1

x^4 + 1 /x^4 = 47

On adding 2 in both sides, we get

x^4 + 1/x^4 + 2 = 47 + 2

=> (x^2) ^2 + ( 1/ x^2)^2 + 2 × (x^2) (1/x^2) = 49

=> ( x^2 + 1 /x^2) ^2 = 49

=> x^2 + 1/x^2 = 7 -----------(1)

Again, on adding 2 in both sides, we get

x^2 + 1/x^2 + 2 = 7 +2

=> ( x + 1/x)^2 = 9

=> x +1/x = 3 --------(2)

Now,

x^3 + 1/x^3 = ( x + 1/x) (x^2 + 1/x^2 - x × 1/x)

= (3) ( 7 - 1)

= 3 × 6

= 18


So,

Option (1) is correct.

Answer 2

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


Add $2(x^{2}* \frac{1}{ x^{2} } )$ i.e. 2  on both sides,

$x^{4} + \frac{1}{ x^{4} } + 2 = 47+2$

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



$x^{2} + \frac{1}{x^{2} } = 7$


Again aad 2 on both sides,


$x^{2} + \frac{1}{ x^{2} } + 2 = 7+2$

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

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

Cube on both sides,


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

$x^{3} + \frac{1}{ x^{3} } + 3(x+ \frac{1}{x} ) = 27$


$x^{3} + \frac{1}{x^{3} } + [3*3] =27$

$x^{3} + \frac{1}{ x^{3} } = 27-9$

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



i hope this will help you


-by ABHAY