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\(1-27x^3\)
\(=1-\left(3x\right)^3\)
\(=\left(1-3x\right)\left(1+3x+9x^2\right)\)
\(---\)
\(x-3^3+27\)
\(=x-27+27=x\)
\(---\)
\(27x^3+27x^2+9x+1\)
\(=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2+1^3\)
\(=\left(3x+1\right)^3\)
\(---\)
\(\dfrac{x^6}{27}-\dfrac{x^4y}{3}+x^2y^2-y^3\) (sửa đề)
\(=\left(\dfrac{x^2}{3}\right)^3-3\cdot\left(\dfrac{x^2}{3}\right)^2\cdot y+3\cdot\dfrac{x^2}{3}\cdot y^2-y^3\)
\(=\left(\dfrac{x^2}{3}-y\right)^3\)
#Ayumu
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a: \(27-27x+9x^2-x^3=\left(3-x\right)^3\)
b: \(125-x^3=\left(5-x\right)\left(25+5x+x^2\right)\)
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\(x^3-9x^2+27x-27\)
=\(x^3-3.x^2.3+3.x.3^3-3^3\)
\(\left(x-3\right)^3\)
Thay x=1, ta có:
(1-3)3=-23=-8
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\(x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
KL:..............
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x3 + 9x2 + 27x + 27 = 0
<=> x2 + 3.3.x2 + 3.32.x + 33 = 0
<=> (x + 3)3 = 0
<=> x + 3 = 0
<=> x = -3
x3 + 9x2 + 27x +27 = 0
=> x3 + 3.x2.3 + 3.x.32 + 33 = 0
=>( x + 3)3 = 0
=> x + 3 = 0
=> x = -3
Vậy x = -3
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\(x^3+9x^2+27x+27=0\)
\(x^3+3\times x^2\times3+3\times x\times3^2+3^3=0\)
\(\left(x+3\right)^3=0\)
\(x+3=0\)
\(x=-3\)