‘Can be solved by most 13-year-olds in China’-5

Algebra Level 3

Whole set
Given real numbers m , n m,n , m ≠ n m\neq n and m 2 = n + 2 , n 2 = m + 2 , m^2=n+2,~~~n^2=m+2, Find the value of m 3 − 2 m n + n 3 . m^3-2mn+n^3.


The answer is -2.

Jeff Giff by Jeff Giff , 13, China

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4 solutions

Jeff Giff
Apr 16, 2021

First, plug m 2 = n + 2 , n 2 = m + 2 m^2=n+2,~~~n^2=m+2 into the polynomial:
m 3 − 2 m n + n 3 = m ⋅ m 2 − 2 m n + n ⋅ n 2 = m ( n + 2 ) − 2 m n + n ( m + 2 ) = m n + 2 m − 2 m n + m n + 2 n = 2 m + 2 n = 2 ( m + n ) . \begin{aligned} m^3-2mn+n^3&=m\cdot m^2-2mn+n\cdot n^2\\ &=m(n+2)-2mn+n(m+2)\\ &=mn+2m-2mn+mn+2n\\ &=2m+2n\\ &=2(m+n). \end{aligned} Now we can try to solve for m + n m+n .
Re-write the equations: { m 2 − n = 2 , n 2 − m = 2. \begin{cases} m^2-n=2,\\ n^2-m=2. \end{cases} ∴ m 2 − n = n 2 − m \therefore m^2-n=n^2-m .
∴ m 2 − n 2 = n − m \therefore m^2-n^2=n-m .
∴ ( m − n ) ( m + n ) = n − m \therefore (m-n)(m+n)=n-m .
Since it is given that m ≠ n m\neq n , m − n ≠ 0 m-n\neq 0 . Therefore we can divide through by m − n m-n .
⇒ m + n = n − m m − n = -1 \Rightarrow m+n=\dfrac{n-m}{m-n}=\color{#D61F06} \fbox{-1} .
Therefore the original polynomial = 2 ( m + n ) = -2 2(m+n)=\colorbox{#CEBB00}{-2} .


6 Helpful 5 Interesting 7 Brilliant 0 Confused

Wow, I never thought about doing it this way. Thanks!

Charley Shi - 1 month, 3 weeks ago

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You’re welcome :) in fact, I never thought I could solve for m m and n n !

Jeff Giff - 1 month, 3 weeks ago

There are some typos when simplifying the polynomial. + 2 m n → − 2 m n +2mn \rightarrow -2mn

Pop Wong - 1 month, 3 weeks ago

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Oh yes! Thanks for telling me!

Jeff Giff - 1 month, 3 weeks ago
Chris Lewis
Apr 16, 2021

Subtracting the equations, m 2 = n + 2 n 2 = m + 2 m 2 − n 2 = n − m m + n = − 1 \begin{aligned} m^2 &=n+2 \\ n^2 &=m+2 \\ m^2-n^2 &=n-m \\ m+n &=-1 \end{aligned}

where we can divide through because m ≠ n m \neq n . If instead we add the equations, m 2 + n 2 = m + n + 4 ( m + n ) 2 − 2 m n = ( m + n ) + 4 1 − 2 m n = 3 m n = − 1 \begin{aligned} m^2+n^2 &=m+n+4 \\ (m+n)^2-2mn &= (m+n)+4 \\ 1-2mn &=3 \\ mn&=-1 \end{aligned}

Now, m 3 − 2 m n + n 3 = ( m + n ) ( m 2 − m n + n 2 ) − 2 m n = ( m + n ) ( ( m + n ) 2 − 3 m n ) − 2 m n = − ( 1 + 3 ) + 2 = − 2 \begin{aligned} m^3-2mn+n^3 &=(m+n) \left(m^2-mn+n^2\right) -2mn \\ &=(m+n)\left((m+n)^2-3mn\right)-2mn \\ &=-(1+3)+2 \\ &=\boxed{-2} \end{aligned}

3 Helpful 1 Interesting 3 Brilliant 0 Confused

Good job! Your solution is nice!

Jeff Giff - 1 month, 3 weeks ago

I realised how you solved for m n mn :) I never wondered how a question like this can have so many different solutions! This is interesting!

Jeff Giff - 1 month, 3 weeks ago
Charley Shi
Apr 16, 2021

Subtract the second equation from the first: m 2 − n 2 = n + 2 − ( m + 2 ) = n − m m^2 - n^2 = n+2 - (m+2) = n-m ( m + n ) ( m − n ) = − ( m − n ) (m+n)(m-n) = -(m-n) Since m ≠ n m\neq n , m − n ≠ 0 m-n\neq 0 so it is valid to divide both sides by m − n m-n . m + n = − 1 ⟶ = m = − 1 − n m+n = -1 \longrightarrow= m = -1 -n Substitute this into equation 2: n 2 = − 1 − n + 2 = 1 − n n^2 = -1-n + 2 = 1-n n 2 + n − 1 = 0 n^2 + n - 1 = 0 n = − 1 ± 5 2 n = \frac{-1 \pm \sqrt{5}}{2} Take the positive root; it doesn't matter actually. m = − 1 − n = − 1 − − 1 + 5 2 = − 1 − 5 2 m = -1-n = -1 - \frac{-1 + \sqrt{5}}{2} = \frac{-1 -\sqrt{5}}{2} Now we get to evaluating m 3 − 2 m n + n 3 m^3 -2mn + n^3 . First calculate the middle term. 2 m n = ( − 1 + 5 ) ) ( − 1 − 5 ) 2 = 1 − 5 2 = − 2 2mn = \frac{(-1+\sqrt{5}))(-1 - \sqrt{5})}{2} = \frac{1 -5}{2} = -2 We can simplify m 3 + n 3 m^3 + n^3 using m = − 1 − n m = -1 - n : m 3 + n 3 = ( − 1 − n ) 3 + n 3 m^3 + n^3 = (-1-n)^3 + n^3 = ( − 1 − n ) ( 1 + 2 n + n 2 ) + n 3 = (-1-n)(1 + 2n + n^2) + n^3 = ( − 1 − 2 n − n 2 − n − 2 n 2 − n 3 ) + n 3 = (-1 -2n -n^2 -n - 2n^2 -n^3) + n^3 = − 1 − 3 n − 3 n 2 = -1 -3n -3n^2 = − 3 ( n 2 + n − 1 ) ⏟ 0 − 4 = − 4 = -3\underbrace{(n^2 + n -1)}_0 -4 = -4 Therefore, m 3 − 2 m n + n 3 = − 4 − ( − 2 ) = − 2 m^3 - 2mn + n^3 = -4 -(-2) = \boxed{-2}

1 Helpful 1 Interesting 2 Brilliant 0 Confused

Good job! But in fact we don’t need to find n n :)

Jeff Giff - 1 month, 3 weeks ago

You would perhaps want to know a quicker way to find ( − 1 − n ) 3 (-1-n)^3 by identity ( a + b ) 3 = a 3 + 3 a 2 b + 3 a b 2 + b 3 (a+b)^3=a^3+3a^2b+3ab^2+b^3 Which is ( − 1 − n ) 3 = ( − 1 ) 3 ( n + 1 ) 3 = − ( n 3 + 3 n 2 + 3 n + 1 ) . (-1-n)^3=(-1)^3(n+1)^3=-(n^3+3n^2+3n+1). :)

Jeff Giff - 1 month, 3 weeks ago

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Yes I know :) Pascal's triangle.

Charley Shi - 1 month, 3 weeks ago
Pop Wong
Apr 16, 2021

Given

m 2 = n + 2 , n 2 = m + 2 m^2 =n+2, n^2= m+2

⟹ m 2 − n 2 = n − m ⟹ ( m + n ) ( m − n ) = n − m ⟹ m + n = − 1 \implies m^2-n^2 = n - m \implies (m+n)(m-n) = n-m \implies m+n = -1

m 3 − 2 m n + n 3 = ( m 3 − m n ) + ( n 3 − m n ) = m ( m 2 − n ) + n ( n 2 − m ) = 2 ( m + n ) = 2 × ( − 1 ) = − 2 m^3-2mn+n^3 = (m^3 - mn) + (n^3-mn) = m(m^2-n) +n(n^2-m) = 2(m+n) = 2\times (-1) = \boxed{-2}

1 Helpful 1 Interesting 1 Brilliant 0 Confused

I see that your solution is the same as mine :) Good job!

Jeff Giff - 1 month, 3 weeks ago

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