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If x4+1x4=14 x^4 + \frac{1}{x^4} = 14x4+x41​=14​, then the value of x3+1x3x^3 + \frac{1}{x^3}x3+x31​​ is A. 363\sqrt{6}36​​​B. 186\fra
Question

If x4+1x4=14 x^4 + \frac{1}{x^4} = 14​, then the value of x3+1x3x^3 + \frac{1}{x^3}​ is
A. 363\sqrt{6}​​
B. 186\frac{18}{\sqrt{6}}​​

C. 9239\sqrt{\frac{2}{3}}​​
D. 323\sqrt{2}​​

A.

A, C, D

B.

B, C, D

C.

A, B, D

D.

A, B, C

Correct option is D

Given:
x4+1x4=14x^4 + \frac{1}{x^4} = 14​​
Let x+1x=yx + \frac{1}{x} = y​​
Formula used:
x2+1x2=y22x4+1x4=(y22)22=y44y2+2x^2 + \frac{1}{x^2} = y^2 - 2 \\x^4 + \frac{1}{x^4} = (y^2 - 2)^2 - 2 = y^4 - 4y^2 + 2​​
Solution:
y44y2+2=14=>y44y212=0y^4 - 4y^2 + 2 = 14 \\\Rightarrow y^4 - 4y^2 - 12 = 0​​
Let z=y2z = y^2​​
Then, z24z12=0z^2 - 4z - 12 = 0​​
=>(z6)(z+2)=0=>z=6\Rightarrow (z - 6)(z + 2) = 0 \\\Rightarrow z = 6 (since y2>0y^2 > 0​)​​
So, y=6y = \sqrt{6}​ or 6-\sqrt{6}​​
Now,
x3+1x3=(x+1x)33(x+1x)=y33yx^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3(x + \frac{1}{x}) \\= y^3 - 3y​​
If y=6,y = \sqrt{6},​​
x3+1x3=(6)336=6636=36x^3 + \frac{1}{x^3} = (\sqrt{6})^3 - 3\sqrt{6} = 6\sqrt{6} - 3\sqrt{6} = 3\sqrt{6}​​
Hence,
x3+1x3=36x^3 + \frac{1}{x^3} = 3\sqrt{6}​​
Correct options are A, B, and C (since they are equivalent forms of 3\sqrt{6}).
Correct answer is (d) A, B, C

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