Question:
If \(\alpha\) and \(\beta\) be two roots of the equation \(x^2-64x+256=0\). Then the value of
\(\left(\dfrac{\alpha^3}{\beta^5}\right)^{1/8}+\left(\dfrac{\beta^3}{\alpha^5}\right)^{1/8}\)
is :
If \(\alpha\) and \(\beta\) be two roots of the equation \(x^2-64x+256=0\). Then the value of
\(\left(\dfrac{\alpha^3}{\beta^5}\right)^{1/8}+\left(\dfrac{\beta^3}{\alpha^5}\right)^{1/8}\)
is :
\(\left(\dfrac{\alpha^3}{\beta^5}\right)^{1/8}+\left(\dfrac{\beta^3}{\alpha^5}\right)^{1/8}\)
is :
Show Hint
Use \(\alpha+\beta=64\) and \(\alpha\beta=256\); rewrite \(\alpha^3/\beta^5=\alpha^3\beta^3/\beta^8\) so that the 8th root is easy.
Updated On: Nov 5, 2026
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The Correct Option is C
Solution and Explanation
From \(x^2-64x+256=0\): \(\alpha+\beta=64,\ \alpha\beta=256\) (both roots are real and positive).
\(\dfrac{\alpha^3}{\beta^5}=\dfrac{\alpha^3\beta^3}{\beta^8}=\dfrac{256^3}{\beta^8}\Rightarrow\left(\dfrac{\alpha^3}{\beta^5}\right)^{1/8}=\dfrac{256^{3/8}}{\beta}=\dfrac{8}{\beta}\) (since \(256^{1/8}=2\)).
Similarly \(\left(\dfrac{\beta^3}{\alpha^5}\right)^{1/8}=\dfrac{8}{\alpha}\).
Sum = \(8\left(\dfrac1\alpha+\dfrac1\beta\right)=\dfrac{8(\alpha+\beta)}{\alpha\beta}=\dfrac{8\times64}{256}=2\).
Answer: (3) 2
\(\dfrac{\alpha^3}{\beta^5}=\dfrac{\alpha^3\beta^3}{\beta^8}=\dfrac{256^3}{\beta^8}\Rightarrow\left(\dfrac{\alpha^3}{\beta^5}\right)^{1/8}=\dfrac{256^{3/8}}{\beta}=\dfrac{8}{\beta}\) (since \(256^{1/8}=2\)).
Similarly \(\left(\dfrac{\beta^3}{\alpha^5}\right)^{1/8}=\dfrac{8}{\alpha}\).
Sum = \(8\left(\dfrac1\alpha+\dfrac1\beta\right)=\dfrac{8(\alpha+\beta)}{\alpha\beta}=\dfrac{8\times64}{256}=2\).
Answer: (3) 2
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