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    If (a+b-c) = 20, and a²+b²+c² = 152, find the value of a³+b³-c³+3abc.
    Question

    If (a+b-c) = 20, and a²+b²+c² = 152, find the value of a³+b³-c³+3abc.

    A.

    640

    B.

    480

    C.

    720

    D.

    560

    Correct option is D

    Given: 
    (a + b - c) = 20, 
    a² + b² + c² = 152 
    Need to find; a³ +  b³ - c³ + 3abc 
    Formula Used: 
    (a+b+c)2=a2+b2+c2+2(ab+bc+ca)a3+b3+c33abc=(a+b+c)(a2+b2+c2abbcca)              =>(a+b+c){(a+b+c)23(ab+bc+ca)}\begin{aligned} & (a+b+c)^2 = a^2 + b^2 +c^2 + 2(ab+bc+ca) \\& a^3 + b^3+c^3 - 3abc = (a+b+c)(a^2 + b^2 +c^2 -ab -bc -ca) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \rArr (a+b+c) \{ (a+b+c)^2 - 3(ab+bc+ca)\} \end{aligned}​​
    Solution:  




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