
`(a + b)^0 = 1`
`(a + b)^1 = a + b`
`(a + b)^ 2 = a^2 + 2ab + b^2`
`(a-b)^ 2 = a^2-2ab + b^2`
`(a + b)^ 3 = a^3 + 3a^2 b + 3ab^2 + b^3`
`(a-b)^ 3 = a^3 -3a^2 b + 3ab^2-b^3`
`(a + b)^4 = a^4 + 4a^3 b + 6a^2 b^2 + 4ab^3 + b^4`
And so on. These formulas are easy to remember because of their symmetry and these are used very frequently in Algebra. We will cover a number of examples that involve these formulas.
`x^2-y^2 = (x + y)(x-y)`
This formula indicates that we can determine the difference of two squares simply my taking a product of the sum and the difference of the variables involved.
`x^3 + y^3 = ( x + y ) ( x^2-xy + y^2 )`
`x^3-y^3 = ( x-y ) ( x^2 + xy + y^2 )`
Using the formula for a perfect square involving the summation sign:
`(a + b) ^2 = a^2 + 2ab + b^2`
That is we need to square the first term, and then square the second term, and then take twice the product of the two terms and then combine the three results.
=> `(x + 5)^2 = x^2 + 2(x)(5) + (5)^2`
=> `(x + 5)^2 = x^2 + 10x + 25`
Using the formula for binomial difference squared:
`(a-b)^ 2 = a^2 -2ab + b^2`
=> `(2x-6)^2 = (2x)^2 -2(2x)(6) + (6)^2`
=> `(2x-6)^2 = 4x^2 -24x + 36`
Using the binomial theorem for cube
`(a-b) ^3 = a^3 -3a^2 b + 3ab^2-b^3`
=> `(2x-2)^3 = (2x)^3-3(2x)^2 (2) + 3(2x)(2)^2-(2)^3`
=> `(2x-2)^3 = 8x3-3(4x2) (2) + 3(2x)(4)-8`
=> `( 2x-2 )^3 = 8x^3-24x^2 + 24x-8`
From the Difference of the Squares Formula, we know that
`x^2-y^2 = (x + y) (x-y)`
=> `(2x + 5) (2x-5) = (2x)^2 -(5)^2`
=> `(2x + 5) (2x-5) = 4x^2 - 25`
We can express the given expression as the sum of cubes form:
`x^3 + 27 = x3 + (3)3`
=> `x^3 + 27 = (x + 3) (x^2-(x)(3) + (3)2)`
=> `x^3 + 27 = (x + 3) (x^2-3x` + 9)
© 2023 iPracticeMath | All Rights Reserved | Terms of Use.