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Algebra Level 3

If a 1 / 3 + b 1 / 3 + c 1 / 3 = 0 a^{1/3} +b^{1/3}+c^{1/3} =0 , which of these is correct?

( a + b + c ) 3 = 27 a b c (a+b+c)^3 =27abc a + b + c = 3 a b c a+b+c=3abc a + b + c = 0 a+b+c=0 a 3 + b 3 + c 3 = 0 a^3+b^3+c^3=0

Ankit Jaiswal by ankit jaiswal , 20

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2 solutions

Mursalin Habib
Apr 13, 2014

This is often useful in problems like these:

If x + y + z = 0 x+y+z=0 , then x 3 + y 3 + z 3 = 3 x y z x^3+y^3+z^3=3xyz

Using that here, we get,

a + b + c = 3 ( a b c ) 1 3 a+b+c=3(abc)^\frac{1}{3}

Then cube:

( a + b + c ) 3 = 27 a b c (a+b+c)^3=27abc .

13 Helpful 0 Interesting 0 Brilliant 0 Confused
Ankit Jaiswal
Apr 12, 2014

a^1/3+b^1/3+c^1/3=0 .:.a^1/3=b^1/3=-c^1/3...............................1 .:.[a^1/3+b^1/3]^3=-c .:.a+b+3[ab]^1/3[a^1/3++b^1/3]=-c .:.a+b+3[ab]^1/3[-c]^1/3=-c [putting a^1/3+b^1/3=-c] .:.a+b-3[abc]^1/3 =-c .:.a+b+c=3[abc]^1/3 .:.[a+b+c]^3=[3[abc]^1/3]^3 .:.[a+b+c]^3=27abc

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Please use LaTeX to make it more informative.

Satvik Golechha - 7 years, 1 month ago

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