Question

If $x^4+\frac{1}{x^4}=194$, find the value of $x^3+\frac{1}{x^3}$.

• A. 64
• B. 52
• C. 40
• D. 38

Answer 1

Answer: (B)

52

Given $:x^4+\frac{1}{x^4}=194$

To find $:x^3+\frac{1}{x^3}=?$

$x^4+\frac{1}{x^4}=194$

$(x^2{)}^2+{\left(\frac{1}{x^2}\right)}^2=194$

${(x^2+\frac{1}{x^2})}^2=194$

${(x^2+\frac{1}{x^2})}^2-2x^2\frac{1}{x^2}=194$ $(a^2+b^2=(a+b{)}^2-2ab)$

${(x^2+\frac{1}{x^2})}^2=194+2$

$x^2+\frac{1}{x^2}=\sqrt{196}$ $x^2+\frac{1}{x^2}=14...\left(1\right)$

${(x+\frac{1}{x})}^2-2.x\frac{1}{x}=14$

$[a^2+b^2=(a+b{)}^2-2ab]$

${(x+\frac{1}{x})}^2=16$

$x+\frac{1}{x}=4...\left(2\right)$

$\therefore x^3+\frac{1}{x^3}=(x+\frac{1}{x})(x^2+\frac{1}{x^2}-1)$

from (1) & (2)

$x^3+\frac{1}{x^3}=4\times (14-1)$

$=4\times 13$

$=52$