What is the factored form of 64g^3+8

8(2g + 1)(4g² - 2g + 1)

Step-by-step explanation:

Given

64g³ + 8 ← factor out 8 from each term

= 8(8g³ + 1) ← 8g³ + 1 is a sum of cubes and factors as

8g³ + 1 = (2g + 1)((2g)² - 2g + 1² ) = (2g + 1)(4g² - 2g + 1)

Hence

64g³ + 8 = 8(2g + 1)(4g² - 2g + 1)

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !