#1**+5 **

Let the number =N

N + N^3 = 68, solve for N

Solve for N:

N^3 + N = 68

Subtract 68 from both sides:

N^3 + N - 68 = 0

The left hand side factors into a product with two terms:

(N - 4) (N^2 + 4 N + 17) = 0

Split into two equations:

N - 4 = 0 or N^2 + 4 N + 17 = 0

Add 4 to both sides:

N = 4 or N^2 + 4 N + 17 = 0

Subtract 17 from both sides:

N = 4 or N^2 + 4 N = -17

Add 4 to both sides:

N = 4 or N^2 + 4 N + 4 = -13

Write the left hand side as a square:

N = 4 or (N + 2)^2 = -13

Take the square root of both sides:

N = 4 or N + 2 = i sqrt(13) or N + 2 = -i sqrt(13)

Subtract 2 from both sides:

N = 4 or N = (0 + 1 i) sqrt(13) - 2 or N + 2 = -i sqrt(13)

Subtract 2 from both sides:

__ Answer: |N = 4__ {or N = (0 + 1 i) sqrt(13) - 2 or N = (0 - i) sqrt(13) - 2} Discard these

Guest Feb 23, 2017

#2**+15 **

Best Answer

**The sum of a number and its cube, is 68. **

**What is the number?**

\(\begin{array}{|rcll|} \hline n+n^3 &=& 68 \\ \underbrace{n}_{=4}\cdot \underbrace{(1+n^2)}_{=17} &=& 4\cdot 17 \\\\ 1+n^2&=&17\\ n^2 &=& 16 \\ n &=& 4 \checkmark \\ \hline \end{array}\)

The number is **4**

heureka
Feb 24, 2017