Question 5

If $$x + \frac{1}{16x} = 3$$ then the value of $$16x^3+\frac{1}{256x^3}$$ is

Solution

$$x + \frac{1}{16x} = 3$$
On multiply by 4,
$$4x + \frac{1}{4x} = 12$$
$$(4x + \frac{1}{4x})^3 = (12)^3$$
$$64x^3 + \frac{1}{64x^3} + 3(4x + \frac{1}{4x}) = 1728$$
$$64x^3 + \frac{1}{64x^3} + 3(12) = 1728$$
$$\frac{1}{4}(64x^3 + \frac{1}{64x^3}) = \frac{1}{4}(1728 - 36)$$
$$(16x^3 + \frac{1}{256x^3}) = 423$$

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SSC CGL Tier-2 12th September 2019 Maths

If $$x + \frac{1}{16x} = 3$$ then the value of $$16x^3+\frac{1}{256x^3}$$ is

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