Question

If


`{ x}^(4) + \frac{1}{x {}^(4) } = 47`
then find the value of

`x {}^(3) + \frac{1}{x {}^(3) }`
​

Answer 1

±18

Explanation:

For this question we need to make use of the following expansions:

`(x + (1)/(x))^(4) \equiv x^(4) + (1)/(x^(4)) + 4\left(x^(2)+ (1)/(x^(2))\right) + 6`

`(x + (1)/(x))^(2) \equiv x^(2) + (1)/(x^(2)) + 2`

`(x + (1)/(x))^(3) \equiv x^(3) + (1)/(x^(3)) + 3(x+(1)/(x))`

So now we can say that

`(x + (1)/(x))^(4) = 47 + 4(x^(2) + (1)/(x^(2))) + 6 \\\\(x + (1)/(x))^(4) = 47 + 4[(x+(1)/(x))^(2)-2]+6 \\\\(x+(1)/(x))^(4) = 45 + 4(x+(1)/(x))^(2) \\\\(x+(1)/(x))^(4) - 4(x+(1)/(x))^(2) - 45 = 0 \\\\(x + (1)/(x))^(2) = 9, (x + (1)/(x))^(2) = -5 \\\\x + (1)/(x) = \pm3`

Hence we get two answers,

`x^(3) + (1)/(x^(3)) = (\pm3)^(3) - 3(\pm3) \\\\x^(3) + (1)/(x^(3)) = 18 \ \text{or} \ x^(3) + (1)/(x^(3)) = -18`