If a+b+c=0, what is the value of a to the power3 + b to the power3 + c to the power3
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Answered by
1
Taking RHS to the identity:
=(a+b+c)(a^2+b^2+c^2 -ab-bc-ac)
=a(a^2+b^2+c^2-ab-bc-ca)+b(a^2+b^2+c^2-ab-bc-ca)+c(a^2+b^2+c^2-ab-bc-ca)
={(a×a^2)+(a×b^2)+(a×c^2)-(a×ab)-(a×bc)-(a×ca)}+{(b×a^2)+(b×b^2)+(b×c^2)-(b×ab)-(b×bc)-(b× ca)}+{(c×a^2)+(c×b^2)+(c×c^2)-(c×ab)-(c×bc)-(c×ca)}
=a^3+b^3+c^3+a^2b+ac^2-ac^2+ab^2-ab^2+bc^2-bc^2+a^2c-a^2c+b^2c-b^2c-abc-abc-abc
=a^3+b^3+c^3-abc-abc-abc
=a^3+b^3+c^3-3abc
=a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
As, (a+b+c)= 0
a^3+b^3+c^3=3abc
=(a+b+c)(a^2+b^2+c^2 -ab-bc-ac)
=a(a^2+b^2+c^2-ab-bc-ca)+b(a^2+b^2+c^2-ab-bc-ca)+c(a^2+b^2+c^2-ab-bc-ca)
={(a×a^2)+(a×b^2)+(a×c^2)-(a×ab)-(a×bc)-(a×ca)}+{(b×a^2)+(b×b^2)+(b×c^2)-(b×ab)-(b×bc)-(b× ca)}+{(c×a^2)+(c×b^2)+(c×c^2)-(c×ab)-(c×bc)-(c×ca)}
=a^3+b^3+c^3+a^2b+ac^2-ac^2+ab^2-ab^2+bc^2-bc^2+a^2c-a^2c+b^2c-b^2c-abc-abc-abc
=a^3+b^3+c^3-abc-abc-abc
=a^3+b^3+c^3-3abc
=a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
As, (a+b+c)= 0
a^3+b^3+c^3=3abc
Answered by
1
Hello,
using the identity :
a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
here if a+b+c=0, we have that:
a³+b³+c³-3abc=0
so
a³+b³+c³=3abc
then the answer will be 3abc
bye :-)
using the identity :
a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
here if a+b+c=0, we have that:
a³+b³+c³-3abc=0
so
a³+b³+c³=3abc
then the answer will be 3abc
bye :-)
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