Question 59

If a+b+c = 6 and ab+bc+ca = 11, then the value of bc(b+c) + ca(c+a) +ab(a+b) +3abc is

Solution

Given : $$(a+b+c)=6$$ and $$ab+bc+ca=11$$ -------(i)

To find : $$bc(b+c)+ca(c+a)+ab(a+b)+3abc$$

= $$(b^2c+bc^2)+(ac^2+a^2c)+(ab^2+a^2b)+3abc$$

= $$(a^2b+a^2c+abc)+(ab^2+b^2c+abc)+(ac^2+bc^2+abc)$$

= $$a(ab+bc+ca)+b(ab+bc+ca)+c(ab+bc+ca)$$

= $$(a+b+c)(ab+bc+ca)$$

Substituting values from equation (i),

=> $$6 \times 11=66$$

=> Ans - (B)

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