Question

If a + b + c = 22 and ab + bc + ca = 91abc, then the value of $\frac{a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)}{abc}$

• A. 2002
• B. 484
• C. 1999
• D. 968

Answer 1

Answer: (C)

1999

Given

$a+b+c=22$ __ (1)

$ab+bc+c9=91abc$ __ (2)

To solve,

$\frac{a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)}{abc}$ ___ (3)

Multiply eq. (1) & (2), we get

$(a+b+c)(ab+bc+ca)=22\times 91abc$

$a^{2}b+abc+a^{2}c+b^{2}a+b^{2}c+abc+abc+c^{2}b+c^{2}a=2002abc$

$a{b}^2+a{c}^2+b{c}^2+b{a}^2+c{a}^2+c{b}^2+3abc=2002abc$

$a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)=2002abc-3abc$

$a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)=abc(2002-3)$

$\frac{a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)}{abc}=1999$