Question

If a^3+b^3+c^3-3abc=p(a^2+b^2+c^2-ab-bc-ca) than find the value of p.​

Answer 1

Answer:

(3abc)/(b+c+a)

Step-by-step explanation:

a^3+b^3+c^3-3abc=p(a^2+b^2+c^2-ab-bc-ca)

p=((a+b+c)(a^2+b^2+c^2)-3*a*b*c)/(a^2+b^2+c^2)(-ab-bc-ca)

  =(a+b+c)(-3abc)/-(ab+bc+ca)

  =(a+b+c)(-3abc)/-(a+b+c)(b+c+a)

  =(3abc)/(b+c+a)

Answer 2

Answer:

a³+b³+c³-3abc

a³+b³+c³-3abc= (a+b+c)(a²+b²+c²-ab-bc-ca)

Step-by-step explanation:

LHS = a³+b³+c³-3abc

= (a³+b³)+c³-3abc

= (a+b)³-3ab(a+b)+c³-3abc

/* By algebraic identity:

x³+y³+3xy(x+y)=(x+y)³

=> x³+y³ = (x+y)³-3xy(x+y) */

= [(a+b)³+c³]-3ab(a+b)-3abc

=[(a+b+c)³-3(a+b)c(a+b+c)]-3ab(a+b+c)

=(a+b+c)[(a+b+c)²-3(a+b)c-3ab]

=(a+b+c)[a²+b²+c²+2ab+2bc+2ca-3ac-3bc-3ab]

=(a+b+c)(a²+b²+c²-ab-bc-ca)

= RHS

Therefore,

a³+b³+c³-3abc

= (a+b+c)(a²+b²+c²-ab-bc-ca)

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