Question

Fatore a seguinte expressão:
a^6 + a^4 + a^2*b^2 + b^4 - b^6​

Answer 1

O exercício solicita a fatoração máxima da seguinte expressão algébrica:

$\mathsf{a^6+a^4+a^2\,b^2+b^4-b^6}$

Para isso, é imprescindível ter conhecimento das duas seguintes propriedades:

$\begin{cases}\mathsf{\!\big(x^p\big)^q=x^{p\cdot q}\,;\,\forall\ x\,\in\,\mathbb{R^{*}},\,\forall\ p,\,q\,\in\,\mathbb{Z}\quad(i)}\\\\ \mathsf{\!\big(x\cdot y\big)^{\!n}=x^n\cdot y^n\,;\,\forall\ x,\,y\,\in\,\mathbb{R^{*}},\,\forall\ n\,\in\,\mathbb{Z}\quad(ii)}\end{cases}$

Para dar continuidade a esta resolução, também é necessário conhecer as quatro identidades algébricas abaixo:

$\begin{cases}\mathsf{yx+x=x\big(y+1\big);\,\forall\ x,\,y\,\in\,\mathbb{C}\quad(iii)}\\\\\mathsf{x^3-y^3=\big(x-y\big)\!\big(x^2+xy+y^2\big);\,\forall\ x,\,y\,\in\,\mathbb{C}\quad(iv)}\\\\ \mathsf{x^2-y^2=\big(x+y\big)\!\big(x-y\big);\,\forall\ x,\,y\,\in\,\mathbb{C}\quad(v)}\\\\ \mathsf{\!\big(x+y\big)^{\!2}=x^2+2xy+y^2\,;\,\forall\ x,\,y\,\in\,\mathbb{C}\quad(vi)}\end{cases}$

Tendo em mente os três resultados fornecidos acima, temos que a expressão algébrica situada no início desta resolução equivaler-se-á:

           

$\mathsf{\quad\, a^6+a^4+a^2\,b^2+b^4-b^6}\\\\\\ \mathsf{=a^6-b^6+a^4+a^2\,b^2+b^4}$

$\mathsf{=\underbrace{\mathsf{a^{2\cdot 3}}}_{(i)}-\,\underbrace{\mathsf{b^{2\cdot 3}}}_{(i)}+\,a^4+a^2\,b^2+b^4}$

$\mathsf{=\underbrace{\mathsf{\big(a^2\big)^{\!3}-\big(b^2\big)^{\!3}}}_{(iv)}+\ a^4+a^2\,b^2+b^4}$

$\mathsf{=\big(a^2-\,b^2\big)\!\Big[\underbrace{\mathsf{\big(a^2\big)^{\!2}}}_{(i)}+\,a^2\,b^2+\underbrace{\mathsf{\big(b^2\big)^{\!2}}}_{(i)}\Big]+a^4+a^2\,b^2+b^4}$

$=\underbrace{\mathsf{\big(a^2-\,b^2\big)\!\big(a^4+a^2\,b^2+b^4\big)+\big(a^4+a^2\,b^2+b^4\big)}}_{\mathsf{(iii)}}$

$\mathsf{=\big(a^4+a^2\,b^2+b^4\big)\!\big[\underbrace{\mathsf{\!\!\big(a^2-\,b^2\big)}}_{(v)}+\,1\big]}$

$\mathsf{=\big(a^4+2a^2\,b^2-a^2\,b^2+b^4\big)\!\big[\!\big(a+b\big)\!\big(a-b\big)+1\big]}$

$\mathsf{=\big(\!\underbrace{\mathsf{a^{2\cdot 2}}}_{(i)}+\ 2a^2\,b^2+\underbrace{\mathsf{b^{2\cdot 2}}}_{(i)}-\underbrace{\,\mathsf{a^2\,b^2}}_{(ii)}\!\big)\!\big[\!\big(a+b\big)\!\big(a-b\big)+1\big]}$

$\mathsf{=\Big[\!\underbrace{\mathsf{\big(a^2\big)^{\!2}+2a^2\,b^2+\big(b^2\big)^{\!2}}}_{(vi)}-\ \big(ab\big)^{\!2}\Big]\!\!\Big[\!\big(a+b\big)\!\big(a-b\big)+1\Big]}$

$\mathsf{=\Big[\!\underbrace{\mathsf{\big(a^2+b^2\big)^{\!2}-\,\big(ab\big)^{\!2}}}_{(v)}\Big]\!\!\Big[\!\big(a+b\big)\!\big(a-b\big)+1\Big]}$

$\mathsf{=\Big[\!\big(a^2+b^2\big)+\big(ab\big)\!\Big]\!\!\Big[\!\big(a^2+b^2\big)-\big(ab\big)\!\Big]\!\!\Big[\!\big(a+b\big)\!\big(a-b\big)+1\Big]}$

$\mathsf{=\big(a^2+ab+b^2\big)\!\big(a^2-ab+b^2\big)\!\big[\!\big(a+b\big)\!\big(a-b\big)+1\big]}\\\\\ \mathsf{\qquad\qquad\qquad\qquad\qquad~~~ \uparrow}\\ \mathsf{\qquad\qquad\qquad Esta\ \'e\ a\ m\'axima\ fatora\,\c \,\!\!c\~ao. }$

Um grande abraço!