Question

sendo m e nas raizes da equaçao x2-10x+1=0, o valor da expressao 1/m3 + 1/n3

Answer 1

Soma dos cubos de 2 termos: $a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$

Quadrado da soma de 2 termos: $(a+b)^{2}=a^{2}+2ab+b^{2}$

Soma de frações (de outra forma):

$\boxed{\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}}$
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$\frac{1}{m^{3}}+\frac{1}{n^{3}}=\frac{n^{3}+m^{3}}{m^{3}n^{3}}\\\\\frac{1}{m^{3}}+\frac{1}{n^{3}}=\frac{(m+n)(n^{2}-mn+m^{2})}{(mn)^{3}}\\\\\frac{1}{m^{3}}+\frac{1}{n^{3}}=\frac{(m+n)(m^{2}+n^{2}-mn)}{(mn)^{3}}$
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$(m+n)^{2}=m^{2}+2mn+n^{2}\\m^{2}+n^{2}=(m+n)^{2}-2mn$
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Substituindo n² + m² por (m + n)² - 2mn:

$\frac{1}{m^{3}}+\frac{1}{n^{3}}=\frac{(m+n)(m^{2}+n^{2}-mn)}{(mn)^{3}}\\\\\frac{1}{m^{3}}+\frac{1}{n^{3}}=\frac{(m+n)([m+n]^{2}-2mn-mn)}{(mn)^{3}}\\\\\frac{1}{m^{3}}+\frac{1}{n^{3}}=\frac{(m+n)([m+n]^{2}-3mn)}{(mn)^{3}}\\\\\boxed{\frac{1}{m^{3}}+\frac{1}{n^{3}}=\frac{S(S^{2}-3P)}{P^{3}}}$

Onde S é a soma das raízes, e P o produto

Calculando S e P:

$x^{2}-10x+1=0\\\\S=-b/a=-(-10)/1=10\\P=c/a=1/1=1$

$\frac{1}{m^{3}}+\frac{1}{n^{3}}=\frac{S(S^{2}-3P)}{P^{3}}\\\\\frac{1}{m^{3}}+\frac{1}{n^{3}}=\frac{10(10^{2}-3.1)}{1^{3}}\\\\\frac{1}{m^{3}}+\frac{1}{n^{3}}=10(100 - 3)\\\\\frac{1}{m^{3}}+\frac{1}{n^{3}}=10(97)\\\\\boxed{\boxed{\frac{1}{m^{3}}+\frac{1}{n^{3}}=970}}$