Without Actually Calculating The Cubes, Find The Value Of:
1. (28)3 + (-15)3 +(-13)3
2. (1/2)3 + (1/3)3 + (5/6)3
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we have to use an identity here
a3+ b3+ c3= ( a + b + c)(a2 +b2+ c2 - ab - bc - ca) + 3abc
i.e.
(28)3 + (-15)3 + (-13)3 = (28+[-15]+[-13])([28]2+[-15]2+[-13] - 28*[-15] - [-15]*[-13] - [-13]*28) + 3[28][-15][-13]
solve this equation and u will get the answer
in the same way part(b) will be done
a3+ b3+ c3= ( a + b + c)(a2 +b2+ c2 - ab - bc - ca) + 3abc
i.e.
(28)3 + (-15)3 + (-13)3 = (28+[-15]+[-13])([28]2+[-15]2+[-13] - 28*[-15] - [-15]*[-13] - [-13]*28) + 3[28][-15][-13]
solve this equation and u will get the answer
in the same way part(b) will be done
Uthandy:
In question one use the formula 10 if,a+b+c=0. a^3+b^3+c^3=3abc
yes u r right ... we n use that in the first part
Yep
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