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Factorise : (a+b)^3+(b+c)^3+(c+a)^3-3(a+...

Factorise : `(a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)`

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To factorize the expression \((a+b)^3 + (b+c)^3 + (c+a)^3 - 3(a+b)(b+c)(c+a)\), we can use a known algebraic identity. ### Step-by-Step Solution: 1. **Recognize the Identity**: The expression can be simplified using the identity: \[ A^3 + B^3 + C^3 - 3ABC = (A + B + C)(A^2 + B^2 + C^2 - AB - AC - BC) ...
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Factorise a^(3)-(b-c)^(3) .

Prove that : (a+b)^(3)+(b+c)^(3)+(c+a)^(3)-3(a+b)(b+c)(c+a)=2(a^(3)+b^(3)+c^(3)-3abc)

Knowledge Check

  • a^3(b-c)^3+b^3(c-a)^3+c^3(a-b)^3

    A
    `3abc`
    B
    `3abc(b-c)(c-a)(a-b)`
    C
    `3(b-c)(c-a)(a-b)`
    D
    none of these

  • If a,b,c are real and distinct nubmer , then the value of ((a -b)^(3) + (b - c)^(3) + (c -a)^(3))/((a - b)(b - c)(c - a)) is

    A
    1
    B
    a b c
    C
    2
    D
    3

    • Similar Questions

    Explore conceptually related problems

    Factorise (a + b + c)^(3) - a^(3) - b^(3) - c^(3) .

    Factorize: (a-3b)^(3)+(3b-c)^(3)+(c-a)^(3)

    ((a-b)^(3)+(b-c)^(2)+(c-a)^(3))/(9(a-b)(b-c)(c-a))=?

    Factorise a^(3) ( b-c) + b^(3) ( c-a) + c^(3) ( a-b) .

    The expression (a-b)^(3)+(b-c)^(3)+(c-a)^(3) can be factorized as (a)(a-b)(b-c)(c-a)(b)3(a-b)(b-c)(c-a)(c)-3(a-b)(b-c)(c-a)(d)(a+b+c)(a^(2)+b^(2)+c^(2)-ab-bc-ca)

    .Factorise: (2a-b-c)^(3)+(2b-c-a)^(3)+(2c-a-b)^(3)

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