Explicación paso a paso:
Formula de binomio al cubo:
(a ± b)³ = a³ ± 3a²b + 3ab² ± b³
Resolvamos:
6.
(2y-1)³ = (2y)³ - 3(2y)²(1) + 3(2y)(1)² - (1)³
(2y-1)³ = 8y³-12y²+6y-1
7.
(-3y+8x)³ = (-3y)³ + 3(-3y)²(8x) + 3(-3y)(8x)² + (8x)³
(-3y+8x)³ = -27y³+216y²x-576yx²+512x³
8.
(3x⁴-6y⁴)³ = (3x⁴)³ - 3(3x⁴)²(6y⁴) + 3(3x⁴)(6y⁴)² - (6y⁴)³
(3x⁴-6y⁴)³ = 27x¹²-162x⁸y⁴+324x⁴y⁸-216y¹²
9.
(9x+2)³ = (9x)³ + 3(9x)²(2) + 3(9x)(2)² + (2)³
(9x+2)³ = 729x³+486x²+108x+8
Formula de binomio al cuadrado:
(a ± b)² = a² ± 2ab + b²
10.
(5x+10y)² = (5x)² + 2(5x)(10y) + (10y)²
(5x+10y)² = 25x²+100xy+100y²
Antes que nada, te voy a dejar la plantilla que se usa en este tipo de operaciones:
Cuando está elevado al cuadrado:
[tex]\bf{\left(a+b\right)^2=a^2+2ab+b^2}[/tex]
[tex]\bf{\left(a-b\right)^2=a^2-2ab+b^2}[/tex]
Cuando está elevado al cubo:
[tex]\bf{\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3}[/tex]
[tex]\bf{\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3}[/tex]
Ahora que ya sabemos la fórmula, resolvemos:
a) [tex]\left(2y-1\right)^3[/tex]
[tex]\left(2y-1\right)^3=\left(2y\right)^3-3\left(2y\right)^2\cdot \:1+3\cdot \:2y\cdot \:1^2-1^3\\\\=\left(2y\right)^3-3\left(2y\right)^2\cdot \:1+3\cdot \:2y\cdot \:1^2-1^3\\\\\boxed{=8y^3-12y^2+6y-1}[/tex]
b) [tex]\left(-3y+8x\right)^3[/tex]
[tex]\left(-3y+8x\right)^3=\left(-3y\right)^3+3\left(-3y\right)^2\cdot \:8x+3\left(-3y\right)\left(8x\right)^2+\left(8x\right)^3\\\\=\left(-3y\right)^3+3\left(-3y\right)^2\cdot \:8x+3\left(-3y\right)\left(8x\right)^2+\left(8x\right)^3\\\\\boxed{=-27y^3+216y^2x-576yx^2+512x^3}[/tex]
c) [tex]\left(3x^4-6y^4\right)^3[/tex]
[tex]\left(3x^4-6y^4\right)^3=\left(3x^4\right)^3-3\left(3x^4\right)^2\cdot \:6y^4+3\cdot \:3x^4\left(6y^4\right)^2-\left(6y^4\right)^3\\\\=\left(3x^4\right)^3-3\left(3x^4\right)^2\cdot \:6y^4+3\cdot \:3x^4\left(6y^4\right)^2-\left(6y^4\right)^3\\\\\boxed{=27x^{12}-162x^8y^4+324x^4y^8-216y^{12}}[/tex]
d) [tex]\left(9x+2\right)^3[/tex]
[tex]\left(9x+2\right)^3=\left(9x\right)^3+3\left(9x\right)^2\cdot \:2+3\cdot \:9x\cdot \:2^2+2^3\\\\=\left(9x\right)^3+3\left(9x\right)^2\cdot \:2+3\cdot \:9x\cdot \:2^2+2^3\\\\\boxed{=729x^3+486x^2+108x+8}[/tex]
e) [tex]\left(5x+10y\right)^2[/tex]
[tex]\left(5x+10y\right)^2=\left(5x\right)^2+2\cdot \:5x\cdot \:10y+\left(10y\right)^2\\\\=\left(5x\right)^2+2\cdot \:5x\cdot \:10y+\left(10y\right)^2\\\\\boxed{=25x^2+100xy+100y^2}[/tex]
MUCHA SUERTE...
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Explicación paso a paso:
Formula de binomio al cubo:
(a ± b)³ = a³ ± 3a²b + 3ab² ± b³
Resolvamos:
6.
(2y-1)³ = (2y)³ - 3(2y)²(1) + 3(2y)(1)² - (1)³
(2y-1)³ = 8y³-12y²+6y-1
7.
(-3y+8x)³ = (-3y)³ + 3(-3y)²(8x) + 3(-3y)(8x)² + (8x)³
(-3y+8x)³ = -27y³+216y²x-576yx²+512x³
8.
(3x⁴-6y⁴)³ = (3x⁴)³ - 3(3x⁴)²(6y⁴) + 3(3x⁴)(6y⁴)² - (6y⁴)³
(3x⁴-6y⁴)³ = 27x¹²-162x⁸y⁴+324x⁴y⁸-216y¹²
9.
(9x+2)³ = (9x)³ + 3(9x)²(2) + 3(9x)(2)² + (2)³
(9x+2)³ = 729x³+486x²+108x+8
Formula de binomio al cuadrado:
(a ± b)² = a² ± 2ab + b²
Resolvamos:
10.
(5x+10y)² = (5x)² + 2(5x)(10y) + (10y)²
(5x+10y)² = 25x²+100xy+100y²
RESOLVER:
Antes que nada, te voy a dejar la plantilla que se usa en este tipo de operaciones:
Cuando está elevado al cuadrado:
[tex]\bf{\left(a+b\right)^2=a^2+2ab+b^2}[/tex]
[tex]\bf{\left(a-b\right)^2=a^2-2ab+b^2}[/tex]
Cuando está elevado al cubo:
[tex]\bf{\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3}[/tex]
[tex]\bf{\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3}[/tex]
Ahora que ya sabemos la fórmula, resolvemos:
a) [tex]\left(2y-1\right)^3[/tex]
[tex]\left(2y-1\right)^3=\left(2y\right)^3-3\left(2y\right)^2\cdot \:1+3\cdot \:2y\cdot \:1^2-1^3\\\\=\left(2y\right)^3-3\left(2y\right)^2\cdot \:1+3\cdot \:2y\cdot \:1^2-1^3\\\\\boxed{=8y^3-12y^2+6y-1}[/tex]
b) [tex]\left(-3y+8x\right)^3[/tex]
[tex]\left(-3y+8x\right)^3=\left(-3y\right)^3+3\left(-3y\right)^2\cdot \:8x+3\left(-3y\right)\left(8x\right)^2+\left(8x\right)^3\\\\=\left(-3y\right)^3+3\left(-3y\right)^2\cdot \:8x+3\left(-3y\right)\left(8x\right)^2+\left(8x\right)^3\\\\\boxed{=-27y^3+216y^2x-576yx^2+512x^3}[/tex]
c) [tex]\left(3x^4-6y^4\right)^3[/tex]
[tex]\left(3x^4-6y^4\right)^3=\left(3x^4\right)^3-3\left(3x^4\right)^2\cdot \:6y^4+3\cdot \:3x^4\left(6y^4\right)^2-\left(6y^4\right)^3\\\\=\left(3x^4\right)^3-3\left(3x^4\right)^2\cdot \:6y^4+3\cdot \:3x^4\left(6y^4\right)^2-\left(6y^4\right)^3\\\\\boxed{=27x^{12}-162x^8y^4+324x^4y^8-216y^{12}}[/tex]
d) [tex]\left(9x+2\right)^3[/tex]
[tex]\left(9x+2\right)^3=\left(9x\right)^3+3\left(9x\right)^2\cdot \:2+3\cdot \:9x\cdot \:2^2+2^3\\\\=\left(9x\right)^3+3\left(9x\right)^2\cdot \:2+3\cdot \:9x\cdot \:2^2+2^3\\\\\boxed{=729x^3+486x^2+108x+8}[/tex]
e) [tex]\left(5x+10y\right)^2[/tex]
[tex]\left(5x+10y\right)^2=\left(5x\right)^2+2\cdot \:5x\cdot \:10y+\left(10y\right)^2\\\\=\left(5x\right)^2+2\cdot \:5x\cdot \:10y+\left(10y\right)^2\\\\\boxed{=25x^2+100xy+100y^2}[/tex]
MUCHA SUERTE...