Question

If a+b+c=5 and ab+bc+ca=10 then prove that a3+b3+c3-3abc= (a+b+c)(a2+b2+c2-ab-bc-ca)

Answer 1

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

Step-by-step explanation:

We are given that a + b + c = 5 and ab + bc + ca = 10.

And we have to prove that: $a^{3} +b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

AS we know that;

$(a+b+c)^{2} =a^{2} +b^{2} +c^{2} +2ab+2bc+2ca$

$(a+b+c)^{2} =a^{2} +b^{2} +c^{2} +2(ab+bc+ca)$

Now, putting the value of (a + b + c) and (ab + bc + ca) as given in the question;

$(5)^{2} =a^{2} +b^{2} +c^{2} +2(10)$

$25 =a^{2} +b^{2} +c^{2} +20$

$a^{2} +b^{2} +c^{2} =25-20$

$a^{2} +b^{2} +c^{2} =5$  ---------------- [Equation 1]

Now, the left-hand side of the equation is;

LHS =  $a^{3} +b^{3}+c^{3}-3abc$

       =  $(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$   {the above equation can be  

                                                                               written as this also}                                

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

       =  $(5)\times (5-(10))$

       =  $5 \times (-5)$  = -25

RHS =  $(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

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

       =  $(5)\times (5-(10))$

       =  $5 \times (-5)$  = -25

Since LHS = RHS, so we have proved that $a^{3} +b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.