Question 68

if x+$$\frac{1}{16x}$$=1, then the value of $$64x{3}+\frac{1}{64x^{3}}$$

Solution

$$x+\ \frac{\ 1}{16x}=1$$

Multiplying by 4 on both sides we get

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

$$\left(4x+\ \frac{\ 1}{4x}\right)^3=4^3$$

$$64x^3\ +\ \ \frac{\ 1}{64x^3}+3.4x.\ \frac{\ 1}{4x}\left(4x+\ \frac{\ 1}{4x}\right)=64$$

$$64x^3\ +\ \ \frac{\ 1}{64x^3}+3\left(4\right)=64$$

$$64x^3\ +\ \ \frac{\ 1}{64x^3}=52$$

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