if a+b+c =5 and ab+bc+ca=10, then prove that a³+b³+c³-3abc= -25.

 a³+b³+c³- 3abc = (a+b+c)*(a²+b²+c²-ab-bc-ca) [It is an identity]

a+b+c= 5 (given).

ab+ bc + ca = 10 (given)

Now concentrate a bit, if you look carefully at the identity, we have got 2 values, we need to find a²+b²+c².

(a+b+c)*[a²+b²+c²- ( ab+bc+ca) ] ( -ab-bc-ca can be written as - (ab+bc+ca, u must know this ofc )

Putting the values we get-

5* ( a²+b²+c²- 10)

We need to find a²+b²+c², AND with values we have been given. If we think a little, we get this :-

(a+b+c)² = a²+b²=c² + 2ab+ 2bc+ 2ca

5² = a²+b²+c²+ 2( ab+bc+ca) [ Taking 2 as common ]

25= a²+b²+c² + 2(10)

25 = a²+b²+c² + 20

Therefore, a²+b²+c²= 5

We got all values! letz put them in the equation we got :-

5* ( 5-10)

= 5* (-5)

= -25.

Proved!

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

a+b+c= 5 (given).

ab+ bc + ca = 10 (given)

5* ( a²+b²+c²- 10)

We need to find a²+b²+c², AND with values we have been given. If we think a little, we get this :-

(a+b+c)²= a²+b²=c²+ 2ab+ 2bc+ 2ca

5²= a²+b²+c²+ 2( ab+bc+ca) [ Taking 2 as common ]

25= a²+b²+c²+ 2(10)

25 = a²+b²+c²+ 20

Therefore, a²+b²+c²= 5

We got all values! put them in the equation we got :-

5* ( 5-10)

= 5* (-5)

= -25.

Hence The Proof!

We got all values! letz put them in the equation we got :-

= -25.

Proved!