solve using identity 25^3-75^3+50^3
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We know that :-
a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
Here , let a =25,b=(-75),c=50
So , a+b+c=25-75+50=0
Substituting a+b+c in identity we get :-
a^3+b^3+c^3-3abc=0
So, a^3+b^3+c^3=3abc
3abc =3*(25)*(-75)*(50)=(-281250)
ANSWER :- 25^3-75^3+50^3=(-281250)
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a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
Here , let a =25,b=(-75),c=50
So , a+b+c=25-75+50=0
Substituting a+b+c in identity we get :-
a^3+b^3+c^3-3abc=0
So, a^3+b^3+c^3=3abc
3abc =3*(25)*(-75)*(50)=(-281250)
ANSWER :- 25^3-75^3+50^3=(-281250)
If you find it helpful please mark it as brainliest....
Answered by
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25^3-75^3+50^3=-281250 is answer.
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