# Thread: sum and differences? 9th grade?

1. ## sum and differences? 9th grade?

Use these patterns to factor polynomials that are sums and differences of cubes.
a^3 + b^3 = (a+b)(a^2-ab+b^2)
a^3-b^3=(a-b)(a^2+ab+b^2)

a)x+8
b)8x^3+27
c)64-y^3
d)2u^3 + 16V^3
e)d^5 + f^3
f) w^6 -1

2. if the sum of two numbers is 6 and the sum of their cubes is 18, what's the sum of their squares?

3. use synthetic division to find (x^2 + 4) / (s - 2i)

Please show work! Thank you so so much!!!!
Give me the answers too.

2. Originally Posted by tecktonikk
Use these patterns to factor polynomials that are sums and differences of cubes.
a^3 + b^3 = (a+b)(a^2-ab+b^2)
a^3-b^3=(a-b)(a^2+ab+b^2)

a)x+8 $\displaystyle \textcolor{red}{(\sqrt[3]{x})^2 + 2^3}$

b)8x^3+27 $\displaystyle \textcolor{red}{(2x)^3 + 3^3}$

c)64-y^3 $\displaystyle \textcolor{red}{4^3 - y^3}$

d)2u^3 + 16V^3 $\displaystyle \textcolor{red}{2[u^3 + (2v)^3]}$

e)d^5 + f^3 $\displaystyle \textcolor{red}{(d^{\frac{5}{3}})^3 + f^3}$

f) w^6 -1 $\displaystyle \textcolor{red}{(w^2)^3 - 1^3}$

2. if the sum of two numbers is 6 and the sum of their cubes is 18, what's the sum of their squares?

$\displaystyle \textcolor{red}{x+y = 6}$

$\displaystyle \textcolor{red}{x^3+y^3 = 18}$

also, note that $\displaystyle \textcolor{red}{(x+y)^2 = 36}$

3. use synthetic division to find (x^2 + 4) / (s - 2i)
Code:
2i] 1 ..... 0 ...... 4

----------------------
all set up ... you do the work .

3. Hello, tecktonikk!

2. If the sum of two numbers is 6 and the sum of their cubes is 18,
what is the sum of their squares?
Let the two numbers be $\displaystyle x$ and $\displaystyle y$.

We have: .$\displaystyle \begin{array}{cccc}x+y &=& 6 & {\color{blue}[1]} \\ x^3+y^3 &=& 18 & {\color{blue}[2]} \end{array}$

Cube equation [1]: .$\displaystyle (x+y)^3 \:=\:6^3 \quad\Rightarrow\quad x^3 + 3x^2y + 3xy^2 + y^3 \:=\:216$

$\displaystyle \text{We have: }\;\underbrace{x^3 + y^3}_{18} + \; 3xy\underbrace{(x+y)}_6 \;=\;216 \quad\Rightarrow\quad 18 \;+\; 18xy \:=\:216 \quad\Rightarrow\quad xy \:=\:11$

Square equation [1]: .$\displaystyle (x+y)^2 \:=\:6^2 \quad\Rightarrow\quad x^2 + 2xy + y^2 \:=\:36$

$\displaystyle \text{We have: }\;x^2 + y^2 + 2\underbrace{(xy)}_{11} \:=\:36 \quad\Rightarrow\quad x^2 + y^2 + 22 \:=\:36 \quad\Rightarrow\quad\boxed{x^2+y^2 \:=\:14}$