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andrew11 [14]
4 years ago
8

How do you solve this problem? (x+3) cubed

Mathematics
2 answers:
Lubov Fominskaja [6]4 years ago
7 0
You would expand becausee it isn't equal to anything

(x+3)^3=(x+3)(x+3)(x+3)
we can use binomial theorem or pascal's triangle
anyway, pascal's triangle is more practicl for this one

so
for the 3rd power
(a+b)^3=1a^3+3a^2b^1+3a^1b^2+1b^3

a=x
b=3

(x+3)^3=1x^3+3x^2(3^1)+3x^1(3)^2+1(3)^3=
x^3+9x^2+27x+27


siniylev [52]4 years ago
3 0
Hi

(x + 3)³
(x+3)*(x+3)*(x+3)
(x+3)*(x²+6x+9)
Now you need to use the distributive property
(x)(x²) + (x)(6x) + (x)(9) +(3)(x²) + (3)(6x) + (3)(9)
Calculate them
x³ + 6x² + 9x + 3x² + 18x + 27
Group the common there
x³ + 6x² + 3x² + 9x + 18x + 27
x³ + 9x² + 27x + 27

I hope that's help !
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Factor completely: (4c - x)^2 - (2c + 3x)^2
azamat

Answer:

Graph for (4*​2.99792e+08-​x)^​2-​(2*​2.99792e+08+​3*​x)^​2

More info

How to solve your problem

(4−)2−1(2+3)2

(4c−x)2−1(2c+3x)2(4c-x)^{2}-1(2c+3x)^{2}(4c−x)2−1(2c+3x)2

Simplify

1

Expand the square

(4−)2−1(2+3)2

(4c−x)2−1(2c+3x)2\left(4c-x\right)^{2}-1(2c+3x)^{2}(4c−x)2−1(2c+3x)2

(4−)(4−)−1(2+3)2

(4c−x)(4c−x)−1(2c+3x)2(4c-x)(4c-x)-1(2c+3x)^{2}(4c−x)(4c−x)−1(2c+3x)2

2

Distribute

(4−)(4−)−1(2+3)2

(4c−x)(4c−x)−1(2c+3x)2{\color{#c92786}{(4c-x)(4c-x)}}-1(2c+3x)^{2}(4c−x)(4c−x)−1(2c+3x)2

4(4−)−(4−)−1(2+3)2

4c(4c−x)−x(4c−x)−1(2c+3x)2{\color{#c92786}{4c(4c-x)-x(4c-x)}}-1(2c+3x)^{2}4c(4c−x)−x(4c−x)−1(2c+3x)2

3

Distribute

4(4−)−(4−)−1(2+3)2

4c(4c−x)−x(4c−x)−1(2c+3x)2{\color{#c92786}{4c(4c-x)}}-x(4c-x)-1(2c+3x)^{2}4c(4c−x)−x(4c−x)−1(2c+3x)2

162−4−(4−)−1(2+3)2

16c2−4cx−x(4c−x)−1(2c+3x)2{\color{#c92786}{16c^{2}-4cx}}-x(4c-x)-1(2c+3x)^{2}16c2−4cx−x(4c−x)−1(2c+3x)2

4

Distribute

162−4−(4−)−1(2+3)2

16c2−4cx−x(4c−x)−1(2c+3x)216c^{2}-4cx{\color{#c92786}{-x(4c-x)}}-1(2c+3x)^{2}16c2−4cx−x(4c−x)−1(2c+3x)2

162−4−4+2−1(2+3)2

16c2−4cx−4cx+x2−1(2c+3x)216c^{2}-4cx{\color{#c92786}{-4cx+x^{2}}}-1(2c+3x)^{2}16c2−4cx−4cx+x2−1(2c+3x)2

5

Combine like terms

162−4−4+2−1(2+3)2

16c2−4cx−4cx+x2−1(2c+3x)216c^{2}{\color{#c92786}{-4cx}}{\color{#c92786}{-4cx}}+x^{2}-1(2c+3x)^{2}16c2−4cx−4cx+x2−1(2c+3x)2

162−8+2−1(2+3)2

16c2−8cx+x2−1(2c+3x)216c^{2}{\color{#c92786}{-8cx}}+x^{2}-1(2c+3x)^{2}16c2−8cx+x2−1(2c+3x)2

6

Expand the square

162−8+2−1(2+3)2

16c2−8cx+x2−1(2c+3x)216c^{2}-8cx+x^{2}-1\left(2c+3x\right)^{2}16c2−8cx+x2−1(2c+3x)2

162−8+2−1(2+3)(2+3)

16c2−8cx+x2−1(2c+3x)(2c+3x)16c^{2}-8cx+x^{2}-1(2c+3x)(2c+3x)16c2−8cx+x2−1(2c+3x)(2c+3x)

7

Distribute

162−8+2−1(2+3)(2+3)

16c2−8cx+x2−1(2c+3x)(2c+3x)16c^{2}-8cx+x^{2}-1{\color{#c92786}{(2c+3x)(2c+3x)}}16c2−8cx+x2−1(2c+3x)(2c+3x)

162−8+2−1(2(2+3)+3(2+3))

16c2−8cx+x2−1(2c(2c+3x)+3x(2c+3x))16c^{2}-8cx+x^{2}-1({\color{#c92786}{2c(2c+3x)+3x(2c+3x)}})16c2−8cx+x2−1(2c(2c+3x)+3x(2c+3x))

8

Distribute

162−8+2−1(2(2+3)+3(2+3))

16c2−8cx+x2−1(2c(2c+3x)+3x(2c+3x))16c^{2}-8cx+x^{2}-1({\color{#c92786}{2c(2c+3x)}}+3x(2c+3x))16c2−8cx+x2−1(2c(2c+3x)+3x(2c+3x))

162−8+2−1(42+6+3(2+3))

16c2−8cx+x2−1(4c2+6cx+3x(2c+3x))16c^{2}-8cx+x^{2}-1({\color{#c92786}{4c^{2}+6cx}}+3x(2c+3x))16c2−8cx+x2−1(4c2+6cx+3x(2c+3x))

9

Distribute

162−8+2−1(42+6+3(2+3))

16c2−8cx+x2−1(4c2+6cx+3x(2c+3x))16c^{2}-8cx+x^{2}-1(4c^{2}+6cx+{\color{#c92786}{3x(2c+3x)}})16c2−8cx+x2−1(4c2+6cx+3x(2c+3x))

162−8+2−1(42+6+6+92)

16c2−8cx+x2−1(4c2+6cx+6cx+9x2)16c^{2}-8cx+x^{2}-1(4c^{2}+6cx+{\color{#c92786}{6cx+9x^{2}}})16c2−8cx+x2−1(4c2+6cx+6cx+9x2)

10

Combine like terms

162−8+2−1(42+6+6+92)

16c2−8cx+x2−1(4c2+6cx+6cx+9x2)16c^{2}-8cx+x^{2}-1(4c^{2}+{\color{#c92786}{6cx}}+{\color{#c92786}{6cx}}+9x^{2})16c2−8cx+x2−1(4c2+6cx+6cx+9x2)

162−8+2−1(42+12+92)

16c2−8cx+x2−1(4c2+12cx+9x2)16c^{2}-8cx+x^{2}-1(4c^{2}+{\color{#c92786}{12cx}}+9x^{2})16c2−8cx+x2−1(4c2+12cx+9x2)

11

Distribute

162−8+2−1(42+12+92)

16c2−8cx+x2−1(4c2+12cx+9x2)16c^{2}-8cx+x^{2}{\color{#c92786}{-1(4c^{2}+12cx+9x^{2})}}16c2−8cx+x2−1(4c2+12cx+9x2)

162−8+2−42−12−92

16c2−8cx+x2−4c2−12cx−9x216c^{2}-8cx+x^{2}{\color{#c92786}{-4c^{2}-12cx-9x^{2}}}16c2−8cx+x2−4c2−12cx−9x2

12

Combine like terms

162−8+2−42−12−92

16c2−8cx+x2−4c2−12cx−9x2{\color{#c92786}{16c^{2}}}-8cx+x^{2}{\color{#c92786}{-4c^{2}}}-12cx-9x^{2}16c2−8cx+x2−4c2−12cx−9x2

122−8+2−12−92

12c2−8cx+x2−12cx−9x2{\color{#c92786}{12c^{2}}}-8cx+x^{2}-12cx-9x^{2}12c2−8cx+x2−12cx−9x2

13

Combine like terms

122−8+2−12−92

12c2−8cx+x2−12cx−9x212c^{2}{\color{#c92786}{-8cx}}+x^{2}{\color{#c92786}{-12cx}}-9x^{2}12c2−8cx+x2−12cx−9x2

122−20+2−92

12c2−20cx+x2−9x212c^{2}{\color{#c92786}{-20cx}}+x^{2}-9x^{2}12c2−20cx+x2−9x2

14

Combine like terms

122−20+2−92

12c2−20cx+x2−9x212c^{2}-20cx+{\color{#c92786}{x^{2}}}{\color{#c92786}{-9x^{2}}}12c2−20cx+x2−9x2

122−20−82

Step-by-step explanation:

7 0
4 years ago
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What is the solution to the equation 2(m - 8) = 4(m + 6) ? m =
ExtremeBDS [4]
The first thing that we want to do is simplify both side of this equation. 

2(m-8).
The first thing we have to do, is multiple 2 by m and -8, also called distributing. So, when we do that, we get
2(m-8)=2*m + 2*-8 = 2m-16

4(m+6)
Now, we do the same thing we did for the first parenthesis, and that's multiple 4 by m and 6. 
4(m+6)=4*m + 4*6 = 4m+24

Now, we need to get m by itself on one side. So, lets bring the equation from the right to the left

2m-16 = 4m + 24     (bring 4m to the other side by subtracting)
2m-16-4m = 24        (bring -16 to the right by adding)
-2m=24+16
-2m=40                    (divide both sides by -2 to get your value for m)
m=40/-2 = -20

So, our answer is m = -20
3 0
3 years ago
Read 2 more answers
What is the circumference of a dinner plate with a radious of 4.5 inches​
____ [38]

Answer:

28.27 inches

Step-by-step explanation:

-Circumference of a circle is calculated using the formula:

C=\pi D\\\\=2\pi r

#Substitute for radius in the formula to solve for C:

C=2\pi r\\\\=2\pi \times 4.5 \\\\=28.27 \ in

Hence, the plate's circumference is 28.27 inches

4 0
3 years ago
-5x - 1 help people
Lady bird [3.3K]

Answer:

5x + 1

Step-by-step explanation:

-5x - 1

to make -5x positive you have to divide both sides by -1

5x + 1

Hope this helps!!!

4 0
3 years ago
Factor this polynomial expression. x^2-16
V125BC [204]

Hi,

We know that:

a² -- b² = (a -- b)(a + b).

With a = x, and b = 4, we have:

x² -- 16 = x² -- 4² = (x -- 4)(x + 4).

As simple as that. Have you understood ?

Green eyes.

6 0
3 years ago
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