Question

find the value of 16 a^4x^3+54ay^3

Answer 1

HI,
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Step  1 :

Equation at the end of step  1  :

 ((16 • ($a^{4}$)) • ($x^{3}$)) + (2•$3^{3}$a$y^{3}$)

 Step  2  :Equation at the end of step  2  : ($2^{4}$$a^{4}$ • $x^{3}$) + (2•$3^{3}$a$y^{3}$)


Step  3  :Step  4  :Pulling out like terms :

 

4.1     Pull out like factors :

  16$a^{4}$$x^{3}$ + 54a$y^{3}$  =   2a • (8$a^{3}$$x^{3}$ + 27$y^{3}$) 

Trying to factor as a Sum of Cubes :

 4.2      Factoring:  8$a^{3}$$x^{3}$ + 27$y^{3}$ 

Theory : A sum of two perfect cubes,  $a^{3}$ + $b^{3}$ can be factored into  :

             (a+b) • ($a^{2}$-ab+$b^{2}$)

Proof  : (a+b) • ($a^{2}$-ab+$b^{2}$) = 

    $a^{3}$-$a^{2}$b+a$b^{2}$+b$a^{2}$-$b^{2}$a+$b^{3}$ =

    $a^{3}$+($a^{2}$b-b$a^{2}$)+(a$b^{2}$-$b^{2}$a)+$b^{3}$=

    $a^{3}$+0+0+$b^{3}$=

    $a^{3}$+$b^{3}$

Check :  8  is the cube of  2 

Check :  27  is the cube of   3 

Check :  $a^{3}$ is the cube of   $a^{1}$

Check :  $x^{3}$ is the cube of   $x^{1}$

Check :  $y^{3}$ is the cube of   $y^{1}$

Factorization is :
             (2ax + 3y)  •  (4x$a^{2}$$x^{2}$ - 6axy + 9$y^{2}$) 

Trying to factor a multi variable polynomial :

 4.3    Factoring    4$a^{2}$$x^{2}$ - 6axy + 9$y^{2}$

Try to factor this multi-variable trinomial using trial and error 

 →Factorization fails

∴Final result : 2a • (2ax + 3y) • (4$a^{2}$$x^{2}$ - 6axy + 9$y^{2}$)
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