a-b=(a^(1/3)-b^(1/3))(a^(2/3)+a^(1/3)b^(1/3)+b^(2/3))
- Thread starter khurram
- Start date
Hello, khurram!
There's less confusion if you use different variables.
\(\displaystyle \text{The difference of cubes: }\:x^3\,-\,y^3\;=\;(x\,-\,y)(x^2\,+\,xy\,+\,y^2)\)
\(\displaystyle \text{Now let }x\,=\,a^{\frac{1}{3}}\text{ and }y\,=\,b^{\frac{1}{3}}\)
\(\displaystyle \text{And we have: }\:\underbrace{\left(a^{\frac{1}{3}}\right)^3\,-\,\left(b^{\frac{1}{3}}\right)^3}\;= \;\left(a^{\frac{1}{3}}\,-\,b^{\frac{1}{3}}\right)\,\left(a^{\frac{2}{3}} \,+\,a^{\frac{1}{3}}b^{\frac{1}{3}}\,+\,b^{\frac{2}{3}}\right)\)
. . . . . . . . . . . . .This is \(\displaystyle a\,-\,b\)
I want to know using which formula they've put L.H.S equal to polynomial at R.H.S?
\(\displaystyle a\,-\,b \;=\;\left(a^{\frac{1}{3}}\,-\,b^{\frac{1}{3}}\right)\,\left(a^{\frac{2}{3}}\,+\,a^{\frac{1}{3}}b^{\frac{1}{3}}\,+\,b^{\frac{2}{3}}\right)\)
If they've used: \(\displaystyle \:a^3\,-\,b^3\:=\a\,-\,b)(a^2\,+\,ab\,+\,b^2)\) . . . then how?
There's less confusion if you use different variables.
\(\displaystyle \text{The difference of cubes: }\:x^3\,-\,y^3\;=\;(x\,-\,y)(x^2\,+\,xy\,+\,y^2)\)
\(\displaystyle \text{Now let }x\,=\,a^{\frac{1}{3}}\text{ and }y\,=\,b^{\frac{1}{3}}\)
\(\displaystyle \text{And we have: }\:\underbrace{\left(a^{\frac{1}{3}}\right)^3\,-\,\left(b^{\frac{1}{3}}\right)^3}\;= \;\left(a^{\frac{1}{3}}\,-\,b^{\frac{1}{3}}\right)\,\left(a^{\frac{2}{3}} \,+\,a^{\frac{1}{3}}b^{\frac{1}{3}}\,+\,b^{\frac{2}{3}}\right)\)
. . . . . . . . . . . . .This is \(\displaystyle a\,-\,b\)