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

b)

`81x^(20)-12x^(15)y^3+(2)/(3)x^(10)y^6-(4)/(243)x^5y^9+(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-(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-(1)/(9)y^3)^4=\sum ^4_(k\mathop=0)\begin{bmatrix}{4} & \\ {k} & {}\end{bmatrix}(3x^5)^(4-k).(-(1)/(9)y^3)^k`

b)

We can now proceed to expand the above.

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

Simplify the above.

`=81x^(20)-12x^(15)y^3+(2)/(3)x^(10)y^6-(4)/(243)x^5y^9+(1)/(6561)y^(12)`

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

`(3x^5-(1)/(9)y^3)^4=81x^(20)-12x^(15)y^3+(2)/(3)x^(10)y^6-(4)/(243)x^5y^9+(1)/(6561)y^(12)`