Algebric identity

Algebra Level pending

a 3 + b 3 + c 3 − 3 a b c ( a − b ) 2 + ( b − c ) 2 + ( c − a ) 2 \large \frac{a^3+b^3+c^3-3abc}{(a-b)^2+(b-c)^2+(c-a)^2}

If a = 25 a=25 , b = 15 b=15 and c = − 10 c=-10 , then find the value of above.

-15 30 15 0 -30

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Plug in the given values of each variable. We have

2 5 3 + 1 5 3 + ( − 10 ) 3 − 3 ( 25 ) ( 15 ) ( − 10 ) ( 25 − 15 ) 2 + ( 15 + 10 ) 2 + ( − 10 + 25 ) 2 = 29250 1950 = \dfrac{25^3+15^3+(-10)^3-3(25)(15)(-10)}{(25-15)^2+(15+10)^2+(-10+25)^2}=\dfrac{29250}{1950}= 15 \boxed{15}

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