Question 45
It is given that $$k^4$$ + $$\frac{1}{k^4}$$ = 47 (where k > 0), then find the value of $$k^3 + \frac{1} {k^3}$$ .
Solution
$$k^4 + \frac{1}{k^4} +2 = 49 => (k^2+\frac{1}{k^2})^2 = 7^2 => k^2 +\frac{1}{k^2} = 7$$.
Similarly, $$k^2 + \frac{1}{k^2} + 2 = 9 => k+\frac{1}{k} = 3.$$
$$(k+1/k)^3 = k^3 + 1/k^3 + 3(k+1/k)=> k^3 +1/k^3 = 27 - 3*3 = 18.$$
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