Question

Find a³ + b³ + c³-3abc/ ab+bc+ca-a²-b²-c² b=-6, c = 10.​

Answer 1

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} )}=

2

1

100−(a

2

+b

2

+c

2

)

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

2

1

(100−50)=25

Now,

= (a³+b³+c³-3abc)

= (a+b+c)

= (a²+b²+c²-ab-bc-ca)

= 10 (50-25)

= 500-250

= 250

Answer 2

Answer:

a3 + b3 + c3 -3abc/ ab+bc+ca-a2 - b2 - c2 = 10.

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