Find 27x3-64y3 if 3x-4y=0 and xy = 12
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Given that,
3x - 4y = 0
and,xy = 12
So,
27x³ - 64y³
= (3x)³- (4y)³
= (3x - 4y)³+ 3xy(3x - 4y)
= (0)³+ 3×12(0)
= 0 +0
= 0
that's it
Hope it helped (ㆁωㆁ*)
3x - 4y = 0
and,xy = 12
So,
27x³ - 64y³
= (3x)³- (4y)³
= (3x - 4y)³+ 3xy(3x - 4y)
= (0)³+ 3×12(0)
= 0 +0
= 0
that's it
Hope it helped (ㆁωㆁ*)
nobel:
hay anyone won't be able to understand that that it is cube
so be careful about this
other wise you won't get any answers
thank god I knew such pattern and solved it for you
ok
bye
ok
bye
bye(*^^)v
(*^^)v
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Chapter = Algebraic expressions and identities
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