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factor(x+y)^3-64
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\(x^3+y^3+3x^3y+3xy^3-4(16)\)
\(x^3+3x^3y+y^3+3xy^3-4(16)\)
(x = y)^3 - 4^3 using difference of 2 cubics a^3 - b^3 = (a - b)(a + ab + b^2) substitute for your answer
oops ( x+ y)^3
x^3(1+3y)+y^3(1+3x)-4(16)
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yup ! @campbell_st
thanks o lot
y :)
for the help
yw :)*
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x^3(1+3y)+y^3(1+3x)-4(16) is the answer na
?
yes!
ok
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