a = b + c = d + e + f
a = b + c (i.e b + c = a)
b + c = d + e + f (i.e d + e + f = b + c)
$$What\ is\ the\ \exp ression\ of\ a(ad+be+df)+be(c+f)+(d+e)f^2+ce(c+f)+f^3?$$
$$=a(ad+be+df)+be(c+f)+(d+e)f^2+ce(c+f)+f^3$$
$$=a(ad+be+df)+be(c+f)+ce(c+f)+\left[df^2+ef^2+ff^2\right]$$
$$=a\left(ad+be+df\right)+\left(bec+bef+cec+cef\right)+\left(d+e+f\right)f^2$$
$$=a\left(ad+be+df\right)+e\left(bc+bf+cc+cf\right)+\left(d+e+f\right)f^2$$
$$=a\left(ad+be+df\right)+e\left(c+f\right)+\left(b+c\right)+\left(d+e+f\right)f^2$$
$$where\ \left(b+c\right)=a\ and\ \left(d+e+f\right)=a$$
$$=a\left(ad+be+df\right)+\left(ec+ef\right)a+af^2$$
$$=aad+bea+dfa+eca+efa+af^2$$
$$=a\left(ad+be+df+ec+ef+f^2\right)$$
$$=a\left(ad+\left(be+ec\right)+\left(df+ef+f^2\right)\right)$$
$$=a\left(ad+e\left(b+c\right)+f\left(d+e+f\right)\right)\ where\ b+c\ =a\ and\ d+e+f=a$$
$$=a\left(ad+ea+fa\right)$$
$$=\left(a\right)\left(a\right)\left(d+e+f\right)\ where\ d+e+f=a$$
$$=\left(a\right)\left(a\right)\left(a\right)=a^3$$
$$Answer=C$$

=>

a(ad + be + df) + be(c + f) + (d + e)f^2 + ce(c + f) + f3
= a(ad + be + df) + be(c + f) + (d + e + f)f^2 + ce(c + f) (combining the terms (d + 3)f^2 and f^3)
= a(ad + be + df) + (b + c)e(c + f) + (d + e + f)f^2 (taking out a common factor of e(c + f) from be(c + f) and ce(c + f))
= a(ad + be + df) + ae(c + f) + af^2 (since a = b + c = d + e + f)
= a(ad + be + df] + a(ce + ef + f^2)
= a[ad + be + df + ce + ef + f^2]
= a[ad + (b + c)e + (d + e + f)f]
= a[ad + ae + af], since a = b + c = d + e + f
= a^2(d + e + f)
= a^3, since a = d + e + f

Therefore, C is the answer.
Answer: C