Question : If $a=299, b=298, c=297$, then the value of $2a^3+2b^3+2c^3-6abc$ is:
Option 1: 5154
Option 2: 5267
Option 3: 5364
Option 4: 5456
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Correct Answer: 5364
Solution :
Given: $a=299, b=298, c=297$
We know that the algebraic identity is $(a^3+b^3+c^3-3abc)=\frac{1}{2} \times(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]$
$2a^3+2b^3+2c^3-6abc$
$=2(a^3+b^3+c^3-3abc)$
$=2\times\frac{1}{2} \times(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]$
$=2\times\frac{1}{2} \times(299+298+297)[(299-298)^2+(298-297)^2+(297-299)^2]$
$=894[1^2+1^2+2^2]=894\times6$
$= 5364$
Hence, the correct answer is 5364.
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