Quadrado da soma multiplicado pela diferença de dois números

Chegou o momento de usar as regras mais avançadas. Multiplique os quadrados das somas pelas diferenças dos mesmos números, usando a regra vista no post anterior.

a)$\underbrace{(ax + by)^{2}}\cdot{\overbrace {(ax – by)}} $

b)$\underbrace {(5 + 3x)^{2}}\cdot{\overbrace{(5 – 3x)}} $

c)$\underbrace {(4n + m^{2})^{2}}\cdot{\overbrace{(4n – m)}} $

d)$\underbrace{(5a + 3b)^{2}}\cdot{\overbrace{(5a – 3b)}}$

e)$\underbrace{(7x + 2y)^{2}}\cdot{\overbrace{(7x -2y)}} $

f)$\underbrace{(10 + 3v)^{2}}\cdot{\overbrace{(10 – 3v)}}$

g)$\underbrace{(px + qy)^{2}}\cdot{\overbrace{(px – qy)}}$

a)$\underbrace{(ax + by)^{2}}\cdot{\overbrace{(ax – by)}}$   $\underbrace{(ax)^3} +\overbrace {(ax)^{2}\cdot(by)} – \underbrace{(ax)\cdot {(by)}^2} -\overbrace {(by)^3}$

$\underbrace{ a^{3}x^{3} }+\overbrace{ a^{2}bx^{2}y} – \underbrace{ab^{2}xy^{2}} -\overbrace {(by)^3}$

$ a^{3}x^{3} + a^2bx^2y – ab^2xy^2 – b^3y^3 $

b)$\underbrace{(5 + 3x)^2}\cdot{\overbrace{(5 – 3x)}}$

$\underbrace{ 5^3} +\overbrace {5^2\cdot{3x}} – \underbrace{5\cdot {(3x)}^2} – \overbrace{(3x)^3}$

$ 125 + 75x – 45x^2 – 27x^3 $

c)$\underbrace{(4n + m^2)^2}\cdot{\overbrace{(4n – m^2)}}$

$\underbrace{(4n)^3} +\overbrace {(4n)^2\cdot(m^2)} -\underbrace{4n\cdot{(m^2)}^2} – \overbrace{(m^2)^3}$

$ 64n^3 + 16n^2m^2 – 4nm^4 – m^6 $

d)$\underbrace{(5a + 3b)^2}\cdot{\overbrace{(5a – 3b)}}$

$\underbrace{(5a)^{3}} +\overbrace{(5a)^{2}\cdot {(3b)}} -\underbrace{ 5a\cdot{(3b)^2}} – \overbrace{(3b)^3}$

$ 125a^3 + 75a^2b – 45ab^2 – 27b^3 $

e)$\underbrace{(7x + 2y)^2}\cdot{\overbrace{(7x -2y)}}$

$\underbrace{(7x)^3} +\overbrace {(7x)^{2}\cdot {(2y)}} -\underbrace{7x\cdot{(2y)^2}} -\overbrace {(2y)^3}$

$ 343x^3 + 98x^2y – 28xy^2 – 8y^3 $

f)$\underbrace{(10 + 3v)^2}\cdot{\overbrace{(10 – 3v)}}$

$\underbrace{(10)^3} +\overbrace{(10)^2\cdot {(3v)}} -\underbrace{10\cdot{(3v)}^2} – \overbrace {(3v)^3}$

$ 1000 + 300v – 90v^2 – 27v^3 $

g)$\underbrace{(px + qy)^2}\cdot{\overbrace{(px – qy)}}$

$\underbrace{(px)^3} +\overbrace {{(px)}^2\cdot {(qy)}} -\underbrace{ px\cdot{(qy)}^2} -\overbrace {(qy)^3}$

$  p^3x^3 + p^2qx^2y – pq^2xy^2 – q^3y^3 $

Uma coleção de exercícios para a prática do conteúdo, pelos leitores/estudantes. Na dúvida, consulte e exponha sua dificuldade para que eu possa ajudá-lo.

h)$\underbrace{(2u + 3v)^2}{\cdot{\overbrace{(2u – 3v)}}}$

i)$\underbrace{(5x + 4y)^2}{\cdot{\overbrace{5x – 4y)}}}$

j)$\underbrace{(3a + 7bc)^2}{\cdot{\overbrace{(3a – 7bc)}}}$

l)$\underbrace{(1 + 9m)^2}{\cdot{\overbrace{(1 – 9m)}}}$

m)$\underbrace{(4p + 6q)^2}{\cdot{\overbrace{(4p – 6q)}}}$

n)$\underbrace{(7x + 3y)^2}{\cdot{\overbrace{7x – 3y)}}}$

Curitiba, 25 de junho de 2018

Décio Adams

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