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
+26367 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
