If x-\(\frac{1}{x}\)=6

than x³+\(\frac{1}{{x}^{3}}\) will be?

74\(\sqrt {10}\)

If x-\(\frac{1}{x}\)=6     [if x-\(\frac{1}{x}\)=k than x+\(\frac{1}{x}\)=\(\sqrt {{k}^{2}+4}\)]

than x+\(\frac{1}{x}\)=\(\sqrt {({6})^{2}+4}\)

x+\(\frac{1}{x}\)=\(\sqrt {40}\) = 2\(\sqrt {10}\)

x³+\(\frac{1}{{x}^{3}}\) = (2\(\sqrt {10}\))³ - 3×2\(\sqrt {10}\)

x³+\(\frac{1}{{x}^{3}}\) = 80\(\sqrt {10}\) - 6\(\sqrt {10}\) = 74\(\sqrt {10}\)