If a+b+c=5;ab+bc+ca=10 find a^3+b^3+c^3-3abc
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Answered by
3
By property
(a+b+c)^2=(a^2+b^2+c^2)+2(ab+bc+ca)
given
a+b+c=5
ab+bc+ca=10
putting value
(5)^2=(a^2+b^2+c^2)+2(10)
25-20=(a^2+b^2+c^2)
5=(a^2+b^2+c^2)
now by this property
a^3 + b^3 + c^3 - 3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)
= (5)(5-(10)
= -25
Answered by
0
Answer:
By property
(a+b+c)^2=(a^2+b^2+c^2)+2(ab+BC+ca)
Given
a+b+c=5
ab+bc+ac=10
Putting value
(5)^2=(a^2+b^2+c^2)+2(10)
25-20=a^2+b^2+c^2
a^2+b^2+c^2=5
Now,by this property:-
a^3+b^3+c^3-3abc=(a+b+c)
(a^2+b^2+c^2-ab-bc-ac)
=(5)(5-(10))
=-25
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