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How do i do this: show that: (x+y+z)2- (x+y)2 = 2(x+y) z+z2

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Work out each side.

the 2x and 2y cancel out leaving

Then the other side

2(x+y) ...... should there be a "+" before the next "z"? or is it multiplied by "2(x+y)"? ... either way, the two are not equivalent

Helpful (2) Fun Thanks for voting Comments (8)
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no there is no "+" before the z...and actually all the "2's on the left side are actually powers..
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powers/ exponents
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ah ... since ask doesn't let ya use superscripts, i'd put a "^" before the 2, like x^2

(x+y+z)(x+y+z) - (x+y)(x+y)
x^2 + xy + xz + xy + y^2 + yz + xz + yz + z^2 - x^2 - xy - xy - y^2
terms that cancel out (x^2, xy, xy, y^2), leaving
xz + yz + xz + z^2, or rearranging terms, z^2 + 2xz + yz

... wait a minute, so, is the right side 2(x+y)^z + z^2 ??
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no the right side is 2(x+y)z+z^2
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so, the other side simplifies to 2xz + 2yz +z^2 ... the other side does too, i just looked over the left side calculation and see that i missed a "yz" term ... but, that's how you do it... just work out each side of the equation
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okaiiii...thanks a lot!! for da help..i understand it now
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also, i recommend not putting a variable after parentheses... i think it's most common to write 2z(x+y) ... but, it's technically right either way :)
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yup!! thank a uuu
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