can anyone answer question 7
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a^2/bc + b^2/ca + c^2/ab
= a^3 + b^3 + c^3 / abc
= 3abc/abc ( a+b+c=0, a^3 + b^3 + c^3 =3abc)
= 3
= a^3 + b^3 + c^3 / abc
= 3abc/abc ( a+b+c=0, a^3 + b^3 + c^3 =3abc)
= 3
Muskaan22:
can you please explain a^3+b^3+c^3/abc?
i mean how did you get it?
a^3+b^3+c^3-3abc=(a+b+c) (a^2+b^2+c^2-ab-bc-ca)
given that a+b+c=0, you put this value in this formula then you got
a^3+b^3+c^3=3abc
ok..i hope it's clear
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sorry , that was by mistakely ,dii I unfollowed u mistakely but then I I followed u again
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thanks for that 4 questions
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