Math

[Guest]

Factor:
(b²+a+a³)²-2(a²-b³-b)(a³+b²+a)+(-b+a²-b³)²+(a⁴-b⁴)²+2(a-b)(a²-b³-b²-b-a³-a)(b²+a²)(b+a).

[Guest]

[ [tex]- a^{4 } + b^{4 } +a^{3 } +b^{3 } - a^{2 } + b^{2 } +ab ][ - a^{4} + b^{4 } + a^{3 } + b^{3 } -a^{2 } + b^{2 } +ab ][/tex]

[Guest]

Solution
Let S=(b²+a+a³)²-2(a²-b³-b)(a³+b²+a)+(-b+a²-b³)²+(a⁴-b⁴)²+2(a-b)(a²-b³-b²-b-a³-a)(b²+a²)(b+a).
Notice that a²-b³-b=-b+a²-b³. Make a substitution: c=a²-b³-b.
Notice that a³+b²+a=b²+a+a³. Make a substitution: d=a³+b²+a.
Notice that a²-b³-b²-b-a³-a=(a²-b³-b)-(a³+b²+a)=c-d.
Let A=2(a-b)(a²-b³-b²-b-a³-a)(b²+a²)(b+a). Simplify A.
The last and precede brackets equal to (a+b) and (a²+b²), respectively. Multiply the first and the last brackets. By the formula of two squares' difference, we will get (a²-b²). At the same way, (a²-b²)(b²+a²)=(a²)²-(b²)². Using the degree's property for variable a and for variable b (the formula of the degree of the degree), we will get: (a²)²-(b²)²=a⁴-b⁴. We will get:
A=2(a⁴-b⁴)(a²-b³-b²-b-a³-a)=2(a⁴-b⁴)(c-d).
Simplify S.
S=d²-2cd+c²+(a⁴-b⁴)²+2(a⁴-b⁴)(c-d).
S=c²-2cd+d²+(a⁴-b⁴)²+2(a⁴-b⁴)(c-d).
Group the first, the second, the third terms and use the formula of the square of the two variables' difference.
S=(c²-2cd+d²)+(a⁴-b⁴)²+2(a⁴-b⁴)(c-d)=(c-d)²+(a⁴-b⁴)²+2(a⁴-b⁴)(c-d).
S=(c-d)²+2(c-d)(a⁴-b⁴)+(a⁴-b⁴)².
Consider the whole sum.
By the formula of the square of the two variables' sum
S=((c-d)+(a⁴-b⁴))².
S=((c-d)+a⁴-b⁴)².
Plug the expression of the (c-d) through a,b into the equation. We will get:
S=((a²-b³-b²-b-a³-a)+a⁴-b⁴)²=(a²-b³-b²-b-a³-a+a⁴-b⁴)².
Explore the polynomial standing in bracket(the base of the square). Let B equal to this polynomial. Perhaps, the factorization of B into the product of the polynomials of non-zero degrees exists.
Group the terms of B pairly in the following way to get the sums (differences) of the degrees with the same indexes:
1)the first with the third;
2)the second with the the fifth;
3)the fourth with the sixth;
4)the seventh with the eighth.
Carry out the minuses for brackets, change the order of the terms of the brackets somewhere, use the formula of the two cubes' sum, identities:
•a²-b²=(a-b)(a+b);
•a⁴-b⁴=(a-b)(a+b)(a²+b²).
B=(a²-b²)+(-b³-a³)+(-b-a)+(a⁴-b⁴)=(a-b)(a+b)-(b³+a³)-(b+a)+(a-b)(a+b)(a²+b²)=(a-b)(a+b)-(a³+b³)-(a+b)+(a-b)(a+b)(a²+b²)=(a-b)(a+b)-(a+b)(a²-ab+b²)-(a+b)+(a-b)(a+b)(a²+b²).
Notice that all the terms have the common divisor (a+b). Carry it out for bracket.
B=(a+b)(a-b-(a²-ab+b²)-1+(a-b)(a²+b²)).
Expand the brackets standing in the largest bracket.
B=(a+b)(a-b-a²+ab-b²-1+a³+ab²-a²b-b³).
The consequence of our exploration is a remarkable factorization:
a²-b³-b²-b-a³-a+a⁴-b⁴=(a+b)(a-b-a²+ab-b²-1+a³+ab²-a²b-b³).
Therefore we will use this important discovery for the our problem's solution.
S=((a+b)(a-b-a²+ab-b²-1+a³+ab²-a²b-b³))².
By the formula of the degree of product
S=(a+b)²(a-b-a²+ab-b²-1+a³+ab²-a²b-b³)².
Answer:(a+b)²(a-b-a²+ab-b²-1+a³+ab²-a²b-b³)².

[Guest]

I found the mistake .
[tex](b^{2 } +a+ a^{3 } )^{2 } -2 (a^{2 } -b^{3 } -b)( a^{3 } + b^{2 } +a) + ( a^{2 } - b^{3 } -b )^{2 }[/tex] +[tex]( a^{4 }- b^{4 } )^{2 }[/tex] +2(a-b)(a+b)[tex]( a^{2 } -b^{2 } -a^{3 } - b^{3 } -a-b)( a^{2 } + b^{2 })[/tex]=

=[tex]( b^{2 } +a+ a^{3 }- a^{2 } + b^{3 }+b )^{2 }[/tex] +[tex]( a^{4 }- b^{4 } )^{2 } +2( a^{2 } - b^2)( a^{2 }+ b^{2 } )( - a^{3 } - b^{3 } +a^{2 } - b^{2 } -a-b)[/tex]=

=[tex]( - a^{3 }- b^{3 } + a^{2 } -b^{2 }-a-b )^{2 } +2(- a^{3 } - b^{3 }+ a^{2 } - b^{2 } -a-b)( a^{4 }- b^{4 }) + ( a^{4 } - b^{4 } )^{2 }[/tex]=

=[tex]( - a^{3 } - b^{3 } +a^{2 }- b^{2 } -a-b+ a^{4 } -b^{4 } )^{2 }[/tex]=

=[tex][ -(a+b)( a^{2 } -ab+b^{2 })+(a-b)(a+b) -(a+b)+(a-b)(a+b)( a^{2 }+ b^{2 }) ]^{2 }[/tex]=

=[tex]( a+b)^{2 }( a^{3 } - b^{3 }- a^{2 }- b^{2 }- a^{2 }b +a b^{2 } +ab+a-b-1 )^{2 }[/tex]