Question

I need help with this practice problem from my trig book. It asks to solve (a) and (b) Please put these ^ separately so I know which is which

Answer 1

a)

$(3x^5-\frac{1}{9}y^3)^4=\sum ^4_{k\mathop=0}\begin{bmatrix}{4} & \\ {k} & {}\end{bmatrix}(3x^5)^{4-k}.(-\frac{1}{9}y)^k$

b)

$81x^{20}-12x^{15}y^3+\frac{2}{3}x^{10}y^6-\frac{4}{243}x^5y^9+\frac{1}{6561}y^{12}$

STEP - BY - STEP EXPLANATION

What to find?

• The sum in summation notation that he used to express the expansion.

,

• The simplified term of the expression.

Given:

$\lbrack3x^5-\frac{1}{9}y^3\rbrack^4$

a)

The binomial theorem formula is given below:

$(a+b)^n=\sum ^n_{k\mathop=0}\begin{bmatrix}{n} & \\ {k} & \end{bmatrix}a^{n-k}b^k$

To obtain the sum in summation that Harold used, simply substitute a=3x⁵ ,

b=-1/9y³ and n=4 into the above formula.

That is;

$(3x^5-\frac{1}{9}y^3)^4=\sum ^4_{k\mathop=0}\begin{bmatrix}{4} & \\ {k} & {}\end{bmatrix}(3x^5)^{4-k}.(-\frac{1}{9}y^3)^k$

b)

We can now proceed to expand the above.

$\begin{gathered} (3x^5-\frac{1}{9}y^3)^4=^4C_0(3x^5)^{4-0}(-\frac{1}{9}y^3)^0+^4C_1(3x^5)^{4-1}(-\frac{1}{9}y^3)^1 \\ +^4C_2(3x^5)^{4-2}(-\frac{1}{9}y^3)^{2^{}}+^4C_3(3x^5)^{4-3}(-\frac{1}{9}x^3)^3+^4C_4(3x^5)^{4-4}(-\frac{1}{9}y^3)^4_{} \end{gathered}$

Simplify the above.

$=81x^{20}-12x^{15}y^3+\frac{2}{3}x^{10}y^6-\frac{4}{243}x^5y^9+\frac{1}{6561}y^{12}$

Hence, the simplified form of the expression is given below:

$(3x^5-\frac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\frac{2}{3}x^{10}y^6-\frac{4}{243}x^5y^9+\frac{1}{6561}y^{12}$