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The sum of a number and its cube, is 68. What is the number?

Guest Feb 23, 2017

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 #2
avatar+18605 
+15

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

 

laugh

heureka  Feb 24, 2017
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2+0 Answers

 #1
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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
avatar+18605 
+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

 

laugh

heureka  Feb 24, 2017

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