\(a,x^3+\frac{1}{27}\)
\(=x^3+\left(\frac{1}{3}\right)^3\)
\(=\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\frac{1}{9}\right)\)
\(b,\left(a+b\right)^3-\left(a-b\right)^3\)
\(=\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=\left(a+b-a+b\right)\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
\(=b^2\left(3a^2+b^2\right)\)
a)x^3+1/27
= x^3 + (1/3)^3
= ( x + 1/3 ) [ x^2 - 1/3 x + (1/3)^2]
]= ( x + 1/3 ) [ x^2 - 1/3 x + 1/9 ]
b)(a+b)^3-(a-b)^3
= a^3 + 3a^2b + 3ab^2 + b^3 - a^3 - 3a^2b + 3ab^2 -+ b^3
( tự rút gọn típ)
Hok tốt nha
Bài này có j khó đâu :v
a) \(x^3+\frac{1}{27}\)
\(=x^3+\left(\frac{1}{3}\right)^3\)
\(=\left(x+\frac{1}{3}\right)\left[x^2-\frac{x}{3}+\left(\frac{1}{3}\right)^2\right]\)
\(=\left(x+\frac{1}{3}\right)\left(x^2-\frac{x}{3}+\frac{1}{3}\right)\)
b) \(\left(a+b\right)^3-\left(a-b\right)^3\)
\(=a^3+3a^2b+3ab^2+b^3-\left(a-b\right)^3\)
\(=a^3+3a^2b+3ab^3+b^3-\left(a^3-3a^2b+3ab^3-b^3\right)\)
\(=6a^2b+3b^2\)
a) \(x^3+\frac{1}{27}=x^3+\left(\frac{1}{3}\right)^3=\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\left(\frac{1}{3}\right)^2\right)\)
\(=\left(x+\frac{1}{3}\right)\left(x^2-\frac{1}{3}x+\frac{1}{9}\right)\)
b) \(\left(a+b\right)^3-\left(a-b\right)^3=\left(\left(a+b\right)-\left(a-b\right)\right)\left(\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right)\)
\(=\left(a+b-a+b\right)\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
=\(=2b\left(3a^2+b^2\right)\)
