Math, asked by Anonymous, 10 months ago

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

Answers

Answered by Anonymous
19

Answer:

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Answered by Anonymous
60

Answer:

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

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