Math, asked by rohith4412, 1 year ago

if ab+bc+ca=10 and a2+b2+c2+44 find a3+b3+c3-3abc

Answers

Answered by humanoid1264
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
gunitmanaktala: According to your logic a+b+c=5 and a2+b2+c2=5
humanoid1264: Pls mark it as brainliest
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Answered by PriyanshSinghTanwar
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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