Bunuel wrote:

Tough and Tricky questions: Exponents.

If \((a+b)^x=a^x + y(a^{(x-1)}*b^{(x-4)}) + z(a^{(x-2)}*b^{(x-3)}) + z(a^{(x-3)}*b^{(x-2)}) + y(a^{(x-4)}*b^{(x-1)}) + b^x\), what is the value of yz?

A. 24

B. 30

C. 36

D. 42

E. 50

Kudos for a correct solution.before solving this question. let's look into the very simple formula of (a+b)^2

(a+b)^2=a^2+b^2+2ab

if we see, the maximum power (or degree) of each term in the expression is 2

so, by using same corollary we can assume that, the highest power of term in the expression (a+b)^x will be x

thus, by considering the second term we have (x-1)+(x-4)=x

or x=5

thus the value of x=5. now the expression given in the question becomes.

\((a+b)^5=a^5 + y(a^{(4)}*b^{(1)}) + z(a^{(3)}*b^{(2)}) + z(a^{(2)}*b^{(3)}) + y(a^{(1)}*b^{(4)}) + b^5\)

also, (a+b)^3= (a+b)(a+b)^2

= (a+b)(a^2+b^2+2ab)

=a^3+ab^2+2(a^2)b+(a^2)b+b^3+2ab^2

=a^3+b^3+3(a^2)b+3ab^2

(a+b)^5= (a+b)^3(a+b)^2

= (a^3+b^3+3(a^2)b+3ab^2)(a^2+b^2+2ab)----------------------1)

also, if we look at the given expression in the question. we will notice that we only have to focus on the co-eff. of terms containing a^4b and a^3b^2

so, let's just focus only on these terms in 1)

=3(a^4)b +2a^4b +a^3b^2+6a^3b^2+3a^3b^2

=5a^4b+10a^3b^2

thus y=5 and z=10

and yz=5(10)=50

p.s. this question can also be solved by using binomial theorem.