Question

11). what is the value of a3 +b3+ c3 - 3abc a=225 b=226 c= 127
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Answer 1

Answer:

$Given:\: a=225, b=226, c=127$

$Using \: formula=a^3+b^3+c^3-3abc =(a+b+c)\frac{1}{2} [(a-b)^2+(b-c)^2+(c-a)^2]$

$a^3+b^3+c^3-3abc=(225+226+227)\frac{1}{2}[(225-226)^2+(226-227)^2+(227-225)^2]$

$a^3+b^3+c^3-3abc=(225+226+227)\frac{1}{2}[(-1)^2+(-1)^2+(2)^2]$

$a^3+b^3+c^3-3abc=(225+226+227)\frac{1}{2}[1+1+4]$

$a^3+b^3+c^3-3abc=(678)\frac{1}{2}[6]$

$a^3+b^3+c^3-3abc=(678)(3)$

$a^3+b^3+c^3-3abc=2078$

Answer 2

Answer:

Here is your answer

hope this helps you