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16 Sep 2014, 01:26
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If \(x\), \(y\) and \(z\) are non-zero numbers, what is the value of \(\frac{x^3+y^3+z^3}{xyz}\)?
(1) \(xyz=-6\)
(2) \(x+y+z=0\)
(1) \(xyz=-6\)
(2) \(x+y+z=0\)
16 Sep 2014, 01:26
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Official Solution:
If \(x\), \(y\) and \(z\) are non-zero numbers, what is the value of \(\frac{x^3+y^3+z^3}{xyz}\)?
(1) \(xyz=-6\).
Infinitely many combinations of \(x\), \(y\), and \(z\) are possible which will give different values of the expression in the stem: try \(x=y=1\) and \(z=-6\), or \(x=1\), \(y=2\), \(z=-3\). Not sufficient.
(2) \(x+y+z=0\).
Rearrange: \(x=-(y+z)\) and substitute this value of \(x\) into the expression in the stem to get:
\(\frac{x^3+y^3+z^3}{xyz}=\)
\(=\frac{-(y+z)^3+y^3+z^3}{xyz}=\)
\(=\frac{-y^3-3y^2z-3yz^2-z^3+y^3+z^3}{xyz}=\)
\(=\frac{-3y^2z-3yz^2}{xyz}=\)
\(=\frac{-3yz(y+z)}{xyz}\).
Since \(x=-(y+z)\) then:
\(\frac{-3yz(y+z)}{xyz}=\)
\(=\frac{-3yz*(-x)}{xyz}=\)
\(=\frac{3xyz}{xyz}=3\).
Sufficient.
Useful property: \((x+y)^3=(x+y)(x^2+2xy+y^2)=x^3+3x^2y+3xy^2+y^3\) and \((x-y)^3=(x-y)(x^2-2xy+y^2)=x^3-3x^2y+3xy^2-y^3\).
Answer: B
If \(x\), \(y\) and \(z\) are non-zero numbers, what is the value of \(\frac{x^3+y^3+z^3}{xyz}\)?
(1) \(xyz=-6\).
Infinitely many combinations of \(x\), \(y\), and \(z\) are possible which will give different values of the expression in the stem: try \(x=y=1\) and \(z=-6\), or \(x=1\), \(y=2\), \(z=-3\). Not sufficient.
(2) \(x+y+z=0\).
Rearrange: \(x=-(y+z)\) and substitute this value of \(x\) into the expression in the stem to get:
\(\frac{x^3+y^3+z^3}{xyz}=\)
\(=\frac{-(y+z)^3+y^3+z^3}{xyz}=\)
\(=\frac{-y^3-3y^2z-3yz^2-z^3+y^3+z^3}{xyz}=\)
\(=\frac{-3y^2z-3yz^2}{xyz}=\)
\(=\frac{-3yz(y+z)}{xyz}\).
Since \(x=-(y+z)\) then:
\(\frac{-3yz(y+z)}{xyz}=\)
\(=\frac{-3yz*(-x)}{xyz}=\)
\(=\frac{3xyz}{xyz}=3\).
Sufficient.
Useful property: \((x+y)^3=(x+y)(x^2+2xy+y^2)=x^3+3x^2y+3xy^2+y^3\) and \((x-y)^3=(x-y)(x^2-2xy+y^2)=x^3-3x^2y+3xy^2-y^3\).
Answer: B
General Discussion
04 Nov 2014, 18:21
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Once it gets to this stage:
−3yz(y+z)
xyz
yz can be canceled out so we're only left with:
−3(y+z)
x
And since x = -(y+z)
3x
x
= 3
−3yz(y+z)
xyz
yz can be canceled out so we're only left with:
−3(y+z)
x
And since x = -(y+z)
3x
x
= 3
