If x^4+1 x^4 =194, then x^3+1 x^3 = - Vidyadip

MCQ

If $\text{x}^4+\frac{1}{\text{x}^4}=194,$ then $\text{x}^3+\frac{1}{\text{x}^3}=$

A
76
B
52
C
64
D
None of these

✓

Answer

52
Solution:
$\text{x}^4+\frac{1}{\text{x}^4}=194$
Now $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=\text{x}^4+\frac{1}{\text{x}^4}+2$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=194+2=196$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=14\ ...(1)$
Now $\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$ $\Big\{\text{x}^2+\frac{1}{\text{x}^2}=14\Big\}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=14+2=16$ [From (1)]
$\Rightarrow\text{x}+\frac{1}{\text{x}}=\sqrt{16}$
$\Rightarrow\text{x}+\frac{1}{\text{x}}=4\ ...(3)$
By identity a³ + b³ = (a + b)(a² + b² - ab)
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=\Big(\text{x}+\frac{1}{\text{x}}\Big)\Big(\text{x}^2+\frac{1}{\text{x}^2}-1\Big)$
$=(4)(14-1)$
$=4\times13$
$=52$
Hence, correct option is (b).

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