It isnC0*A^n*b^0 + nC1*A^(n-1)*b^1 + ... + nCr*A^(n-r)*b^r + ... + nCn*A^0*b^n
where nCr = n!/[r!*(n-r)!]
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(ax + b)3 = a3x3 + 3a2bx2 + 3ab2x + b3
give me something to answer and ill answer it ASAP.. :D but here is my example 2 2 (a+b) (a-b) =a -b
Depends on the kind of binomials. Case 1: If both binomials have different terms, then use the distribution property. Each term of one binomial will multiply both terms of the other binomial. After distribution, combine like terms, and it's done. Case 2: If both binomials have exactly the same terms, then work as in the 1st case, or use the formula for suaring a binomial, (a ± b)2 = a2 ± 2ab + b2. Case 3: If both binomials have terms that only differ in sign, then work as in the 1st case, or use the formula for the sum and the difference of the two terms, (a - b)(a + b) = a2 - b2.
Depends on the kind of binomials. Case 1: If both binomials have different terms, then use the distribution property. Each term of one binomial will multiply both terms of the other binomial. After distribution, combine like terms, and it's done. Case 2: If both binomials have exactly the same terms, then work as in the 1st case, or use the formula for suaring a binomial, (a ± b)2 = a2 ± 2ab + b2. Case 3: If both binomials have terms that only differ in sign, then work as in the 1st case, or use the formula for the sum and the difference of the two terms, (a - b)(a + b) = a2 - b2.
2pi/3 radian or equivalent 120 degree