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Find the value of (x+y+z)^3

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by Aman verma » Wed Jun 02, 2010 1:27 am
Q: If x^1/3 + y^1/3 + z^1/3 = 0 , then the value of ( x+y+z )^3 is :

a) 27

b) 27xyz

c) 81

d)(xyz)^3

e)xyz
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by Rahul@gurome » Wed Jun 02, 2010 1:42 am
Let x^1/3 = a, y^1/3 = b, z^1/3 = c
The question then becomes, if (a+b+c) = 0, then what is the value of (a^3+b^3+c^3)^3.
We have a+b = -c.
Or (a+b)^3 = (-c)^3 = -c^3.
Or a^3+b^3+3*a*(b^2)+3*(a^2)*b = -c^3
Or a^3+b^3+c^3 = -{ 3*a*(b^2)+3*(a^2)*b } = -3*a*b*(a+b) = -3*a*b*(-c) = 3abc.
So (a^3+b^3+c^3)^3 = (3abc)^3 = 27(a^3)(b^3)(c^3) = 27xyz.
The correct answer is (b).
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by jube » Wed Jun 02, 2010 2:02 am
b

Choosing 8, 1 & -1 as x, y & z gets b
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by selango » Wed Jun 02, 2010 4:16 am
Try plugging using,

x=8

y=-1

z=-1

(x+y+z)^3=6^3

27*8=27*xyz

Option B
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by Thouraya » Wed Jun 02, 2010 4:26 am
Sorry, can you please explain further where did u get the numbers u plugged?
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by selango » Wed Jun 02, 2010 4:37 am
Try to plugin the numbers so that the below equation is satisified.

x^3+y^3+z^3=0

Let x=8,y=-1,z=-1.The above equation value will be zero.

Now substitute these values in (x+y+z)^3.
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by liferocks » Wed Jun 02, 2010 4:40 am
here we can use a formula,
if a+b+c=0
a^3+b^3+c^3=3abc
here a=x^1/3 b=y^1/3 and c=z^1/3
so x+y+z=3(xyz)^1/3
or ( x+y+z )^3=27xyz

Ans option D
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by Aman verma » Wed Jun 02, 2010 9:11 am
Awesome !!! OAB. Thanks to all ! Love the approach adopted by Rahul. Others approach are also good. An innovative & time saving approach is always welcome .
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by Thouraya » Tue Apr 12, 2011 12:59 am
What's the easiest way to solve this kind of problem? Thanks
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