Calculate the Following Values
\[ x+\frac{1}{x}=3 \]
Find:
\[ x^2+\frac{1}{x^2},\quad x^3+\frac{1}{x^3},\quad x^4+\frac{1}{x^4} \]
Solution:
Using identity:
\[ \left(x+\frac{1}{x}\right)^2 = x^2+\frac{1}{x^2}+2 \]
\[ (3)^2 = x^2+\frac{1}{x^2}+2 \]
\[ 9 = x^2+\frac{1}{x^2}+2 \]
\[ x^2+\frac{1}{x^2} = 7 \]
Now using identity:
\[ a^3+b^3=(a+b)^3-3ab(a+b) \]
\[ x^3+\frac{1}{x^3} = \left(x+\frac{1}{x}\right)^3 -3\left(x\cdot\frac{1}{x}\right)\left(x+\frac{1}{x}\right) \]
\[ = (3)^3-3(1)(3) \]
\[ = 27-9 \]
\[ = 18 \]
Now,
\[ \left(x^2+\frac{1}{x^2}\right)^2 = x^4+\frac{1}{x^4}+2 \]
\[ (7)^2 = x^4+\frac{1}{x^4}+2 \]
\[ 49 = x^4+\frac{1}{x^4}+2 \]
\[ x^4+\frac{1}{x^4} = 47 \]