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Algebra Level 2

99 9 3 − 12 3 3 − 87 6 3 999^3 - 123^3 - 876^3

is equal to:

999 + 123 + 876 999 + 123 + 876 3 × 999 × 123 × 876 3 \times 999 \times 123 \times 876 99 9 2 + 12 3 2 + 87 6 2 999^2 + 123 ^2 + 876 ^ 2 999 × 123 × 876 999 \times 123 \times 876

Chung Kevin by Chung Kevin , 45, Australia

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2 solutions

Sravanth C.
Jul 9, 2015

Let: a = 999 a=999 , b = 123 b=123 and c = 876 c=876 .

Here we have, a 3 − b 3 − ( a − b ) 3 = a 3 − b 3 − ( a 3 − b 3 − 3 a b ( a − b ) = a 3 − b 3 − a 3 + b 3 + 3 a b ( a − b ) = 3 a b ( a − b ) a^3-b^3-(a-b)^3\\ =a^3-b^3-(a^3-b^3-3ab(a-b)\\ =a^3-b^3-a^3+b^3+3ab(a-b)\\=3ab(a-b)

Now, substituting the values of a a , b b and c c : 3 × 999 × 123 × ( 999 − 123 ) = 3 × 999 × 123 × 876 3\times 999\times 123\times(999-123)\\=\boxed{3×999×123×876}

7 Helpful 0 Interesting 1 Brilliant 0 Confused

Moderator note:

Good use of the algebraic identity.

The other way of approaching this is see that if a + b + c = 0 a + b + c = 0 , then

a 3 + b 3 + c 3 = 3 a b c a^3 + b^3 + c^3 = 3abc

Nihar Mahajan
Jul 10, 2015

I did by letting a = 999 , b = − 123 , c = − 876 a=999 \ , \ b=-123 \ , \ c = -876 .

So we have a + b + c = 0 a+b+c=0 .We have familiar identity :

a 3 + b 3 + c 3 = 3 a b c = 3 × 999 × ( − 123 ) × ( − 876 ) = 3 × 999 × ( 123 ) × ( 876 ) a^3+b^3+c^3 = 3abc = 3 \times 999 \times (-123) \times (-876) \\ = 3 \times 999 \times (123) \times (876)

3 Helpful 0 Interesting 0 Brilliant 0 Confused

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