Question

1. $(x^{2}-2y^{3})(x^{2}-2y^{3})$
2. $(x^{2}-2y^{3})^{2}$
3. $(1 - 3x)^{2}$
4. $(a + b)^{2}$
5. $(y + 2)(y^{2}-3y + 4)$
6. $(a - b)(a^{2}+ab + b^{2})$

Explanation:

Step1: Expand using square formula

For problem 13/14: $(x^2-2y^3)^2=(x^2)^2-2(x^2)(2y^3)+(2y^3)^2$
$=x^4 - 4x^2y^3 + 4y^6$

Step2: Expand $(1-3x)^2$

$(1-3x)^2=1^2-2(1)(3x)+(3x)^2$
$=1 - 6x + 9x^2$

Step3: Expand $(a+b)^2$

$(a+b)^2=a^2+2ab+b^2$

Step4: Distribute $(y+2)$ to trinomial

$(y+2)(y^2-3y+4)=y(y^2-3y+4)+2(y^2-3y+4)$
$=y^3-3y^2+4y+2y^2-6y+8$
$=y^3 - y^2 - 2y + 8$

Step5: Expand difference of cubes

$(a-b)(a^2+ab+b^2)=a(a^2+ab+b^2)-b(a^2+ab+b^2)$
$=a^3+a^2b+ab^2-a^2b-ab^2-b^3$
$=a^3 - b^3$

Answer:

1. $x^4 - 4x^2y^3 + 4y^6$
2. $x^4 - 4x^2y^3 + 4y^6$
3. $9x^2 - 6x + 1$
4. $a^2 + 2ab + b^2$
5. $y^3 - y^2 - 2y + 8$
6. $a^3 - b^3$