CUET Preparation Today
If $det. (A-B) ≠0, A^4=B^4, C^3A=C^3B$ and $B^3A = A^3B$, then find the value of $det. (A^3+ B^3 + C^3)$. |
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$(A^3+ B^3 + C^3)(A-B)$ $=A^4-A^3B+B^3A-B^4+C^3A-C^3B=O$ $∴det.((A^3+ B^3 + C^3)(A-B))=0$ $⇒det.(A^3+ B^3 + C^3)=0$ as $det.(A-B)≠0$ |
