Question

If $3x+\frac{1}{3x}=3;find:\left(i\right)9x^2+\frac{1}{9x^2}\left(ii\right)27x^3+\frac{1}{27x^3}$

Answer 1

(i) $3x+\frac{1}{3x}=3\Rightarrow {(3x+\frac{1}{3x})}^2=(3{)}^2\Rightarrow {(3x+\frac{1}{3x})}^2=(3{)}^2\Rightarrow (3x{)}^2+{\left(\frac{1}{3x}\right)}^2+2\times 3x\times \frac{1}{3x}=9\Rightarrow 9x^2+\frac{1}{9x^2}+2=9\Rightarrow 9x^2+\frac{1}{9x^2}=9-2\Rightarrow 9x^2+\frac{1}{9x^2}=7$.

(ii) $3x+\frac{1}{3x}=3\Rightarrow {(3x+\frac{1}{3x})}^3=(3{)}^3\Rightarrow (3x{)}^3+{\left(\frac{1}{3x}\right)}^3+3\times 3x\times (3x+\frac{1}{3x})=27\Rightarrow 27x^3+\frac{1}{27x^3}+3\left(3\right)=27\Rightarrow 27x^3+\frac{1}{27x^3}=27-9\Rightarrow 27x^3+\frac{1}{27x^3}=18$.