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  • If a + b + c = 5 and ab + bc + ca = 10, then prove that a^{3} + b^{3} + c^{3} –3abc = – 25

If a + b + c = 5 and ab + bc + ca = 10, then prove that a3 + b3 + c3 –3abc = – 25

Answers (1)

Given: (a + b + c) = 5, ab + bc + ca = 10

To Prove: a3 + b3 + c3 –3abc = – 25

We know that

(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)\\ (5)^{2}=a^{2}+b^{2}+c^{2}+2(10)\\ 25=a^{2}+b^{2}+c^{2}+20\\ 25-20=a^{2}+b^{2}+c^{2}\\ 5=a^{2}+b^{2}+c^{2}

Now,

 a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

Putting the values, we get: 

a^{3}+b^{3}+c^{3}-3abc=5\left [ 5-(ab+bc+ca) \right ]\\ =5(5-10)\\ =5(-5)\\ =-25\\=R.H.S

Hence proved.                                              

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