Exclude-m-and-n-from-the-equalities-a-m-n-b-3-m-3-n-3-c-5-m-5-n-5-

Question Number 109544 by 1549442205PVT last updated on 25/Aug/20

Exclude m and n from the equalities :

a = m + n, b^{3} = m^{3} + n^{3}, c^{5} = m^{5} + n^{5}

Commented by Her_Majesty last updated on 24/Aug/20

o n l y 2 e q u a l i t i e s a r e n e c e s s a r y t o g e t m a n d

n \Rightarrow {i t}^{'} s a n e a s y t a s k

l e t m = x - y \land n = x + y

a = 2 x \Rightarrow x = \frac{a}{2}

b^{3} = 3 {a y}^{2} + \frac{a^{3}}{4} \Rightarrow y^{2} = \frac{4 b^{3} - a^{3}}{12 a}

c^{5} = \frac{5 b^{6} + 5 a^{3} b^{3} - a^{6}}{9 a}

Answered by ajfour last updated on 24/Aug/20

m^{3} + n^{3} = (m + n) {{(m + n)}^{2} - 3 m n}

b^{3} = a (a^{2} - 3 m n)

m^{5} + n^{5} = (m^{3} + n^{3}) (m^{2} + n^{2})

- m^{2} n^{2} (m + n)

\Rightarrow

c^{5} = b^{3} {a^{2} - \frac{2}{3} (a^{2} - \frac{b^{3}}{a})} - \frac{a}{9} {(a^{2} - \frac{b^{3}}{a})}^{2}

Commented by 1549442205PVT last updated on 25/Aug/20

Thank both Sirs

Answered by 1549442205PVT last updated on 25/Aug/20

From the hypothesis we have

a^{3} = {(m + n)}^{3} = m^{3} + n^{3} + 3 mn (m + n) =

b^{3} + 3 mna \Rightarrow mn = \frac{a^{3} - b^{3}}{3 a} (1)

a^{5} = {(m + n)}^{5} = m^{5} + n^{5} + 5 mn (m^{3} + n^{3})

+ 10 {(mn)}^{2} (m + n) = c^{5} + {5 mnb}^{3} +

+ 10 mna (2) . Replace (1) into (2) we get

\boldsymbol a^{5} - \boldsymbol c^{5} = 5 \boldsymbol b^{3} . \frac{a^{3} - b^{3}}{3 a} + 10 a . {(\frac{a^{3} - b^{3}}{3 a})}^{2}

= \frac{a^{3} - b^{3}}{3 a} ({5 b}^{3} + 10 a . \frac{a^{3} - b^{3}}{3 a})

= \frac{a^{3} - b^{3}}{3 a} . \frac{{10 a}^{3} + {5 b}^{3}}{3} . Finally, we obtain

9 \boldsymbol a (\boldsymbol a^{5} - \boldsymbol c^{5}) = 5 (\boldsymbol a^{3} - \boldsymbol b^{3}) (2 \boldsymbol a^{3} + \boldsymbol b^{3})