if a+b+c=5 and ab+ bc+ca=10 prove that a×a×a+b×b×b+c×c×c-3abc=-25
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Given:
a+b+c=5 and ab+bc+ca=10
In a³+b³+c³-3abc=-25
Taking LHS,
=a³+b³+c³-3abc
=(a+b+c)(a²+b²+c²-ab-bc-ca) -------------------1
As we know that
a²+b²+c²+2(ab+bc+ca)=(a+b+c)²
a²+b²+c²=(a+b+c)²-2(ab+bc+ca)
substituting the given values
a²+b²+c²=(5)²-2(10) ⇒25-20 ⇒5 ------------------2
From 1 and 2
=(a+b+c)(a²+b²+c²-(ab+bc+ca))
=(5)(5-10)
=5(-5) ⇒-25 ⇒RHS
:) Hope this helps!!!!!!!!
a+b+c=5 and ab+bc+ca=10
In a³+b³+c³-3abc=-25
Taking LHS,
=a³+b³+c³-3abc
=(a+b+c)(a²+b²+c²-ab-bc-ca) -------------------1
As we know that
a²+b²+c²+2(ab+bc+ca)=(a+b+c)²
a²+b²+c²=(a+b+c)²-2(ab+bc+ca)
substituting the given values
a²+b²+c²=(5)²-2(10) ⇒25-20 ⇒5 ------------------2
From 1 and 2
=(a+b+c)(a²+b²+c²-(ab+bc+ca))
=(5)(5-10)
=5(-5) ⇒-25 ⇒RHS
:) Hope this helps!!!!!!!!
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