InterviewSolution
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InterviewSolution
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Prove that: `|((b+c)^2,a^2,a^2),(b^2,(c+a)^2,b^2),(c^2,c^2,(a+b)^2)|=2a b c(a+b+c)^3` |
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Answer» Let the given determinant be `Delta`. Then. `Delta =|{:((b+c)^(2), " "a^(2), a^(2)),(b^(2),(c+a)^(2), b^(2)),(c^(2)," "c^(2) ,(a+b^(2))):}|` `=|{:((b+c)^(2)-a^(2), " "0, " "a^(2)),(" "0,(c+a)^(2)-b^(2), " "b^(2)),(c^(2)-(a+b)^(2),c^(2)-(a+b)^(2),(a+b^(2))):}| [C_(1) to C_(1)-C_(3) " and "C_(2) to C_(2) -C_(3)]` `=|{:((a+b+c)(b+c-a), " "0, " "a^(2)),(" "0,(a+b+c)(c+a-b), " "b^(2)),((a+b+c)(c-a-b),(a+b+c)(c-a-b),(a+b^(2))):}|` `=(a+b+c)^(2) *|{:((b+c-a), " "0, " "a^(2)),(" "0,c+a-b, " "b^(2)),(c-a-b,c-a-b,(a+b^(2))):}| ["taking (a + b+c) common from"C_(1) "and"C_(2) "both"]` `=(a+b+c)^(2) *|{:((b+c-a), " "0, " "a^(2)),(" "0,c+a-b, " "b^(2)),(-2b,-2a, 2ab):}| [R_(3) to R_(3) -(R_(1) + R_(2))]` `=(a+b+c)^(2)[(b+c-a){(c+a-b) *2b+2ab^(2)} + a^(2){0 +2b(c+a-b)}]` `=(a+b+c)^(2)[(b+c-a)*2ab{(c+a-b+b)} + 2a^(2)b(c+a-b)]` `=2ab(a+b+c)^(2){(b+c-a)(c+a)+a(c+a-b)}` `=2ab(a+b+c)^(2) * {bc+ab+c^(2)+ac-ac-a^(2)+ac+a^(2)-ab}` `=2ab(a+b+c)^(2){bc+c^(2)+ac} =2abc(a+b+c)^(3)` Hence, `Delta = 2abc(a+b+c)^(3)` |
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