Question

Show that (a-b)³ + (b-c)³ + (c-a)³ = 3(a-b) (b-c) (c-a)​

Answer 1

ǫᴜᴇsᴛɪᴏɴ →

Show that (a-b)³ + (b-c)³ + (c-a)³ = 3(a-b) (b-c) (c-a)

ᴀɴsᴡᴇʀ →

• Answer in attachment.

• hope it's help you.

Answer 2

Answer:

(a2−b2)3+(b2−c2)3+(c2−a2)3(a−b)3+(b−c)3+(c−a)3

Let in numerator

x=(a2−b2),y=(b2−c2),z=(c2−a2)

⟹x+y+z=0(1)

We know, (x+y+z)3=x3+y3+z3+3(x+y+z)(xy+yz+zx)−3xyz

0=x3+y3+z3+3(0)(xy+yz+zx)−3xyz

x3+y3+z3=3xyz(2)

Let in denominator,

p=(a−b),q=(b−c),r=(c−a)

⟹p+q+r=0

⟹

p3+q3+r3=3pqr(3)

(a2−b2)3+(b2−c2)3+(c2−a2)3(a−b)3+(b−c)3+(c−a)3=3xyz3pqr

=(a2−b2)(b2−c2)(c2−a2)(a−b)(b−c)(c−a)

=(a−b)(b−c)(c−a)(a+b)(b+c)(c+a)(a−b)(b−c)(c−a)

=(a+b)(b+c)(c+a)

We know that if x+y+z = 0 then a^3+b^3+c^3=3ABC

Now (a^2-b^2)+(b^2-c^2)+(c^2-a^2) = 0

So (a^2-b^2)^3 + (b^2-c^2)^3 + (c^2-a^2)^3

= 3(a^2-b^2)(b^2-c^2)(c^2-a^2)

= 3(a+b)(a-b)(b+c)(b-c)(c+a)(c-a) ……..(1)

Similarly (a-b)+(b-c)+(c-a) = 0

=> (a-b)^3+(b-c)^3+(c-a)^3 =3(a-b)(b-c)(c-a)….(2)

From (1) and (2)

{(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3} /

{(a-b)^3+(b-c)^3+(c-a)^3}

=3(a+b)(a-b)(b+c)(b-c)(c+a)(c-a) / 3(a-b)(b-c)(c-a)

= (a+b)(b+c)(c+a).

(a2−b2)3+(b2−c2)3+(c2−a2)3(a−b)3+(b−c)3+(c−a)3We know that if A+B+C=0ThenA3+B3+C3=3ABCAnd this is true for numerator and denominator3(a+b)(a−b)(b+c)(b−c)(c−a)(c+a)3(a−b)(b−c)(c−a)=:(a+b)(b+c)(c+a)