Question

Find the value of a³ + b³ + c³ - 3abc when (a+b+c) = 10 and a² + b² + c² = 50.​

Answer 1

$\huge\bf\underline\red{Solution:}$

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Answer 2

a+b+c = 10

(a+b+c)² = a²+b²+c²+2(ab+bc+ca) = 100

ab+bc+ac =$\frac {1}{2} \: {100 - ( {a}^{2} + {b}^{2} + {c}^{2} )}$

= $\frac{1}{2} (100 - 50) = 25$

Now,

= (a³+b³+c³-3abc)

= (a+b+c)

= (a²+b²+c²-ab-bc-ca)

= 10 (50-25)

= 500-250

= 250