Question:
If \( a\cos^3\alpha + 3a\cos\alpha\sin^2\alpha = m \) and \( a\sin^3\alpha + 3a\cos^2\alpha\sin\alpha = n \), then \( (m+n)^{2/3} + (m-n)^{2/3} \) is equal to
If \( a\cos^3\alpha + 3a\cos\alpha\sin^2\alpha = m \) and \( a\sin^3\alpha + 3a\cos^2\alpha\sin\alpha = n \), then \( (m+n)^{2/3} + (m-n)^{2/3} \) is equal to
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Use $(x+y)^2+(x-y)^2 = 2(x^2+y^2)$ trick.
Updated On: Apr 23, 2026
- $2a^3$
- $2a^{1/3}$
- $2a^{2/3}$
- $2a^3$
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The Correct Option is C
Solution and Explanation
Concept:
Use identity of $(x+y)^3$ :
Step 1: \[ m = a(\cos^3\alpha + 3\cos\alpha\sin^2\alpha) \]
Step 2: \[ m = a(\cos\alpha(\cos^2\alpha + 3\sin^2\alpha)) \]
Step 3: Recognize: \[ \cos^3\alpha + 3\cos\alpha\sin^2\alpha = (\cos\alpha + \sin\alpha)^3 - \sin^3\alpha \]
Step 4: Similarly simplify $m+n$: \[ m+n = a(\cos\alpha + \sin\alpha)^3 \]
Step 5: Hence: \[ (m+n)^{2/3} = a^{2/3}(\cos\alpha + \sin\alpha)^2 \]
Step 6: Similarly: \[ (m-n)^{2/3} = a^{2/3}(\cos\alpha - \sin\alpha)^2 \]
Step 7: Add: \[ = a^{2/3}[(\cos+\sin)^2 + (\cos-\sin)^2] \] \[ = a^{2/3}[2(\cos^2 + \sin^2)] = 2a^{2/3} \] Conclusion:
Answer = $2a^{2/3}$
Step 1: \[ m = a(\cos^3\alpha + 3\cos\alpha\sin^2\alpha) \]
Step 2: \[ m = a(\cos\alpha(\cos^2\alpha + 3\sin^2\alpha)) \]
Step 3: Recognize: \[ \cos^3\alpha + 3\cos\alpha\sin^2\alpha = (\cos\alpha + \sin\alpha)^3 - \sin^3\alpha \]
Step 4: Similarly simplify $m+n$: \[ m+n = a(\cos\alpha + \sin\alpha)^3 \]
Step 5: Hence: \[ (m+n)^{2/3} = a^{2/3}(\cos\alpha + \sin\alpha)^2 \]
Step 6: Similarly: \[ (m-n)^{2/3} = a^{2/3}(\cos\alpha - \sin\alpha)^2 \]
Step 7: Add: \[ = a^{2/3}[(\cos+\sin)^2 + (\cos-\sin)^2] \] \[ = a^{2/3}[2(\cos^2 + \sin^2)] = 2a^{2/3} \] Conclusion:
Answer = $2a^{2/3}$
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