Find the Value
\[ x^2+\frac{1}{x^2}=98 \]
Find:
\[ x^3+\frac{1}{x^3} \]
Solution:
Using identity:
\[ \left(x+\frac{1}{x}\right)^2 = x^2+\frac{1}{x^2}+2 \]
\[ \left(x+\frac{1}{x}\right)^2 = 98+2 \]
\[ \left(x+\frac{1}{x}\right)^2 = 100 \]
\[ x+\frac{1}{x} = 10 \]
Now using identity:
\[ a^3+b^3=(a+b)^3-3ab(a+b) \]
Here,
\[ a=x,\quad b=\frac{1}{x},\quad ab=1 \]
\[ x^3+\frac{1}{x^3} = \left(x+\frac{1}{x}\right)^3 -3\left(x+\frac{1}{x}\right) \]
\[ = (10)^3-3(10) \]
\[ = 1000-30 \]
\[ =970 \]