bhedamohammed
bhedamohammed bhedamohammed
  • 23-04-2020
  • Mathematics
contestada

If a + b + c.= 5 and ab + bc +ca = 10 , then prove that a3+ b3+ c3-3abc = -25
ANSWER WITH WORKING THX

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LammettHash
LammettHash LammettHash
  • 23-04-2020

Start with

[tex](a+b+c)^3=a^3+b^3+c^3+3(a^2b+a^2c+ab^2+ac^2+b^2c+bc^2)+6abc[/tex]

Given that [tex]a+b+c=5[/tex], we know [tex](a+b+c)^3=125[/tex].

Also, [tex]ab+bc+ca=10[/tex], so

[tex]a^2b+a^2c+ab^2+ac^2+b^2c+bc^2=a(ab+ac)+b(ab+bc)+c(ac+bc)[/tex]

[tex]=a(10-bc)+b(10-ca)+c(10-ab)[/tex]

[tex]=10(a+b+c)-3abc[/tex]

[tex]=50-3abc[/tex]

So we have

[tex]125=a^3+b^3+c^3+3(50-3abc)+6abc[/tex]

[tex]125=a^3+b^3+c^3+150-9abc+6abc[/tex]

[tex]\implies -25=a^3+b^3+c^3-3abc[/tex]

QED

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