Question : If $\left(y-\frac{1}{y}\right)=4$, find the value of $\left(y^6+\frac{1}{y^6}\right)$.
Option 1: 5774
Option 2: 4096
Option 3: 5776
Option 4: 5778
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Correct Answer: 5778
Solution : Given: $\left(y-\frac{1}{y}\right)=4$ Cubing both sides, we get, ⇒ $\left(y-\frac{1}{y}\right)^{3}=4^{3}$ ⇒ $y^{3}-\frac{1}{y^{3}}-3×y×\frac{1}{y}(y-\frac{1}{y})=64$ ⇒ $y^{3}-\frac{1}{y^{3}}-3(4)=64$ ⇒ $y^{3}-\frac{1}{y^{3}}=64+12=76$ Again, squaring both sides, we get $(y^{3}-\frac{1}{y^{3}})^2=76^2=5776$ ⇒ $\left(y^3)^2 + (\frac{1}{y^3}\right)^2-2×y^3×\frac{1}{y^3}=5776$ ⇒ $\left(y^6+\frac{1}{y^6}\right)=5776+2=5778$ Hence, the correct answer is 5778.
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Question : If $\left(y+\frac{1}{y}\right)=4$, then find the value of $\left(y^6+\frac{1}{y^6}\right)$.
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