(102)^3+(98)^3
Solve it using cube identities
Please answer with explanation
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let 102 = a and 98 = b
a^3 + b^3 = (a+b)(a^2 -ab + b^2)
so here it is
(102+98)(102^2+102×98+98^2)
200 (10404+9996+9604)
200×30040
= 6000800
HOPE IT HELPS! ! !
a^3 + b^3 = (a+b)(a^2 -ab + b^2)
so here it is
(102+98)(102^2+102×98+98^2)
200 (10404+9996+9604)
200×30040
= 6000800
HOPE IT HELPS! ! !
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