what is a cubed minus a squared b plus a b squared minus b cubed factored out?
444043
a^3 - a^2*b + a*b^2 - b^3 =\n= (a^3 + a*b^2) - (a^2*b + b^3) =\n= a(a^2 + b^2) - b(a^2 + b^2) =\n= (a - b)*(a^2 + b^2)
1027973
the 3rd one is correct, but I get -(3+b)/b+2 and 2 for the first 2
4519
(4/3)*pi*r^3\n\n[four thirds pi r cubed]\n\ngood luck!
87985
The expression was not set equal to anything.\n\nIf you're doing simple factoring in algebra class, the \nsimplest factors would be (b+4) (b+3).\n\nIf the expression was set to equal 0, solve for b, then your answer set would include -4 and -3.\n\nHope this helps!
930643
factor:\n\n(2x^2+x-6)/(2x+4)=\n\n(2x-3)(x+2) / 2(x+2)\n\nthe (x+2)s cancel out leaving\n\n(2x-3)/2
914755
your equation must be like x^2-3x-6 and your trying to factor this\n\n2 numbers say a and b must multiply to make -6 and add to make -3 so a*b=-3 and a+b=-6\n but i can't think on any that fits this criteria so you must use the quadratic equation\n\nx= [-b +-squareroot(b^2-4ac)]/2a\nwhere a=1 b=-3 and c=-6\nx= 3+-squareroot(9-4(1)(-6))/2\n\nx=(3+squareroot(33))/2 and x= 3-squareroot(33)/2\n\nin decimals x=4.3723 and x=-1.3723
767438
b1=(-b+ (b^2-4ac)^1/2 )/2a=(-8+(64-48)^1/2)/6=(-8+4)/6\n    =-4/6=-2/3\nb2=(-b- (b^2-4ac)^1/2 )/2a=(-8-4)/6=-12/6=-2
475059
4\ncubed root of 8 is 2. squared is 4. \nOr 8 squared is 64, cubed root of that is 4.
130182
The difference of two sqares can always be factored in the form of a^2 – b^2 = (a+b)(a-b)\n\nSo it follows that a^2 – 9 = (a+3)(a-3) So now you have\n\n(a+3)(a-3)(4a)\n-----------------\n(8a)(a-3)\n\nCancel out the (a-3) in the top and bottom, and also the a’s in the 4a & 8a.  Also, the 4 cancels with the 8 and leaves a 2 in the denominator.  So your simplified answer is\n\n(a+3)/2
703982
1.\n3x + 6/((x + 1)(x + 2)) Factor 3 out of the top.\n[3(x + 2)]/[(x + 1)(x + 2)] x+2's cancel\n3/(x + 1) <-- This is simplified.\n\nI don't have the energy to do the other ones right now.  :/
652728
a=b=i^(1/3)\ni=-1\nab+a^2b+ab^2 = [i^(1/3)*i^(1/3)] + [i^(2/3)*i^(1/3)] + [i^(1/3)*i^(2/3)]\n= [i^(2/3)] + i + i\n= -2+(-1)^(2/3)
945505
{(x^2+5x+4)/(x+3)^2}/(x+1/(x^2 -9))\nFactor everything that you can factor.\nx^2 + 5x + 4 = (x+1)(x+4)\nx^2 - 9 = (x+3)(x-3)\nYou have\n{(x+1)(x+4)/(x+3)^2} * {(x+3)(x-3)/(x+1)}\nYou can cancel the (x+3)'s and the (x+1)'s leaving you with\n(x+4)(x-3)/(x+3)
875476
(a + b - c)(a + b + c)\n((a+b)+c)((a+b)-c)\n(a+b)^2-c^2\na^2+2ab+b^2-c^2
282094
x^3 - 9x \nx appears in both terms so you can factor out an x\nx(x^2 - 9)  \nx^2 - 9 is a difference of squares. \nUse the formula a^2 - b^2 = (a+b)(a-b)\nFor x^2 - 9 you have x^2 - 3^2, so this factors into\n(x+3)(x-3)\nHence,\nx^3 - 9x\n= x(x+3)(x-3)
378246
a)\n-1[3x(2x+3)-2(6)]\n= -1[6x^2 + 9x -12]\n= -6x^2 - 9x + 12\n\nb) Factors of -4 that add up to 0 are -2 and 2 because\n(-2)(2) = -4 and -2+2 = 0\nThus, x^2-4 = (x-2)(x+2)\n\nc) x^3 + 1 is a sum of two squares, namely x and 1\nThe formula is \na^3 + b^3 = (a+b)(a^2 - ab + b^2)\nLet a = x and b = 1\nx^3 + 1 = (x+1)(x^2 - x + 1)\n\nNote: Since it just says simplify part a), you do not need to factor. (If the problem wanted it to be factored, then it would say to simplify and factor.)
959197
I have factored it please refer to.\n\nFrom this solution can be found
498372
I think u are talkin abt this method....\n\n\nif u hav a cubic like  x^3 + 2x^2 - 2x - 1 = 0\n\nto find out whether (x-1) is a factor....\n\nfirst write the coeff of the eq in this way n write the factor ( here it iz 1 since x-1=0 or x=1) on the left hand side Like this\n\n{one}  1     2     -2     -1\n--------  0     1      3      1\n_______________________________\n{sum}  1     3      1      0\n\n\nHere what i hav done is write the 0 under the coeff of x^3\nnow add that coeff n the zero\nu get 1...\nnow multiply this one wid the number written on the left hand side(here it is also one)\nwite what u get under the coeff of x^2....\nas above add the new no and the coeff of x^2\nin a similar way multiply one and the sum u got n write it under the coeff of x...\nadd the no with coeff of x n then write it below....\nfinally u multiply the sum wid one and write it under the constant term.... n then add...\nif u get 0 as the sum then whatever no u hav taken in da LHS is on of the root.... here we get 0 in the last case... so (x-1) is a fctor.....\n\nnow this eq can b written as \n\n    (x-1) * (1*x^2 + 3 * x + 1) = 0\nhere 1 , 3 and 1 of the quadratic are the sum u got in the above method......