If a + b + c = 6 and $$a^3 + b^3 + c^3 - 3abc$$ = 126, then ab + bc + ca is equal to:

Given that,

a + b + c = 6

$$a^3 + b^3 + c^3 - 3abc = 126$$

We know that $$(a+b+c)^2=(a^2+b^2+c^2+2ab+2bc+2ca)$$

$$\Rightarrow 6^2 -(2ab+2bc+2ca)=a^2+b^2+c^2$$

$$\Rightarrow 36 -(2ab+2bc+2ca)=a^2+b^2+c^2 ----------------(i)$$

Now,

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

substituting the values in equation (i) and (ii), 

$$\Rightarrow 126= (6)(36-2ab-bc-ca-ab-bc-ca)$$

$$\Rightarrow 126= (6)(36-3ab-3bc-3ca)$$

$$\Rightarrow 126= 6\times36-18ab-18bc-18ca$$

$$\Rightarrow 18ab-18bc-18ca=216-126$$

$$\Rightarrow ab+bc+ca=\dfrac{90}{18}$$

$$\Rightarrow ab+bc+ca=5$$