If ab=21, then $$\frac{(a+b)^2}{(a-b)^2}=\frac{25}{4}$$ then $$a^2+b^2+3ab$$ the value is 

Given : $$\frac{(a+b)^2}{(a-b)^2}=\frac{25}{4}$$

=> $$\frac{a^2+b^2+2ab}{a^2+b^2-2ab}=\frac{25}{4}$$

Let $$(a^2+b^2)=x$$ and $$ab=21$$

=> $$\frac{x+42}{x-42}=\frac{25}{4}$$

=> $$4x+(42 \times 4)=25x-(42 \times 25)$$

=> $$25x-4x=(42 \times 4)+(42 \times 25)$$

=> $$21x=42 \times (4+25)$$

=> $$x=\frac{42 \times 29}{21}$$

=> $$x=2 \times 29=58$$

$$\therefore$$ $$a^2+b^2+3ab$$

= $$58+3(21)$$

= $$58+63=121$$

=> Ans - (B)