if ab+bc+ca=10 and a2+b2+c2+44 find a3+b3+c3-3abc
Answers
Answered by
38
We know ,
a³ + b³ + c³ -3abc = (a + b + c )(a² + b² + c² -ab -bc-ca)
now ,
a + b + c = 5
ab + bc + ca = 10
(a + b + c)² = a² + b² + c² +2(ab + bc+ca)
(5)² -2×10 = a² + b² + c²
a² + b² + c² =5
hence ,
a³ + b³ +c³ -3abc = ( a + b + c )(a² + b² + c² -ab- bc-ca)
=( 5)( 5 - 10) = 5 × (-5) = -25
hence proved
gunitmanaktala:
ugh
Answered by
45
Answer:
Step-by-step explanation:
(a+b+c)^2=a2+b2+c2+2(ab+ bc+ca)
(a+b+c)^2=44+20
(a+b+c)^2=64
a+b+c=8
Now a3+ b3 +c3-3abc=(a2+b2+c2-(ab+bc+ac)(a+b+c)
Therefore putting all values we have a3+b3+c3-3abc=8(44-10)
=272
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