Question

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

A
64
B
52
C
40
D
38

Solution

## The correct option is C 52Given $$: x^{4}+\dfrac{1}{x^{4}} = 194$$To find $$: x^{3}+\dfrac{1}{x^{3}} = ?$$$$x^{4}+\dfrac{1}{x^{4}} = 194$$$$(x^{2})^{2}+\left ( \dfrac{1}{x^{2}} \right )^{2} = 194$$$$\left ( x^{2}+\dfrac{1}{x^{2}} \right )^{2} = 194$$$$\left ( x^{2}+\dfrac{1}{x^2} \right )^{2}-2x^{2}\dfrac{1}{x^{2}} = 194$$    $$(a^{2}+b^{2} = (a+b)^{2}-2ab)$$$$\left ( x^{2}+\dfrac{1}{x^{2}} \right )^{2} = 194+2$$$$x^{2}+\dfrac{1}{x^{2}} = \sqrt{196}$$   $$\boxed {x^{2}+\dfrac{1}{x^{2}} = 14}...(1)$$$$\left ( x+\dfrac{1}{x} \right ) ^{2}-2.x\dfrac{1}{x} = 14$$$$[a^{2}+b^{2} = (a+b)^{2}-2ab]$$$$\left ( x+\dfrac{1}{x} \right )^{2} = 16$$$$\boxed {x +\dfrac{1}{x} = 4}...(2)$$$$\therefore x^{3}+\dfrac{1}{x^{3}} = \left ( x+\dfrac{1}{x} \right )\left ( x^{2}+\dfrac{1}{x^{2}}-1 \right )$$from (1) & (2)$$x^{3}+\dfrac{1}{x^{3}} = 4\times (14-1)$$$$= 4\times 13$$$$= \boxed {52}$$ Maths

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