P1 math as level cie .

the coefficient of x^3 in the expansion of (1-3x)^6 + (1+ax)^5 is 100 . Find the value of the constant a.

I got for x^3
10a^3 x^3 + (-180a^2 x^3) + -540x^3 + 675a x^3

But why does the markscheme just write the 10a^3x^3 and - 540x^3 and not the other numbers that I got along with those that also have x^3 ?

the coefficient of x^3 in the expansion of (1-3x)^6 + (1+ax)^5 is 100 . Find the value of the constant a.

I got for x^3
10a^3 x^3 + (-180a^2 x^3) + -540x^3 + 675a x^3

But why does the markscheme just write the 10a^3x^3 and - 540x^3 and not the other numbers that I got along with those that also have x^3 ?

There is no (-180a^2 x^3) or 675a x^3. I'm not sure where you're getting these terms from. In a binomial expansion, there will be only one term of each power - (1-3x)^6 gives 6C3 * (-3)^3 x^3 = -540x^3, and (1+ax)^5 gives 5C3 * a^3 x^3 = 10a^3 x^3.

The coefficient of $x^3$ in the expansion of $(1-3x)^6$ is $\displaystyle \binom{6}{3}(-3)^3 = -540$

The coefficient of $x^3$ in the expansion of $(1+ax)^5$ is $\displaystyle \binom{5}{3}(a)^3 = 10a^3$


Hence the coefficient of $x^3$ in the expansion of $(1-3x)^6 + (1+ax)^5$ is $-540+10a^3$

Why don’t u write the other x^3 that you can get between those two expansions ? Why just those two numbers ?

You need to add the expansions, not multiply them

You need to add the expansions, not multiply them

It says in the expansion , so u have to expand and then add , u have to expand to get the x^3 , then add

It's (1-3x)^6 (1 ax)^5

Not (1-3x)^6(1 ax)^5.