Question

3. expand the binomial, (x^2-y^2)^3 4. find the fourth term of (a-b^2)^5

Answer 1

3. (x^2-y^2)^3 = x^6 - 3(x^4)(y^2) + 3(x^2)(y^4) - y^6

4. fourth term is -10(a^2)(b^6)

Answer 2

Answer:

$(x^2+(-y^2))^3=x^6+3\cdot x^4\cdot (-y^2)+3\cdot x^2\cdot y^4+3\cdot (-y^6)$

$T_{3+1}=10\cdot a^{2}\cdot -b^6$  is the fourth term.

Step-by-step explanation:

We have formula for binomial expansion:

$(a+b)^n=^{n}C_0\cdot a^{n-0}\cdot b^0+^{n}C_1\cdot a^{n-1}\cdot b^1+^{n}C_1\cdot a^{n-2}\cdot b^2+---+^{n}C_n\cdot a^0\cdot b^n$

Here a=x^2, b= -y^2 and n=3

Substituting the values in the formula we get and using $^{n}C_r=\frac{n!}{r!\cdot (n-r)!}$

$(x^2+(-y^2))^3=^{3}C_0\cdot (x^2)^{3-0}\cdot (-y^2)^0+^{3}C_1\cdot (x^2)^{3-1}\cdot (-y^2)^1+^{3}C_2\cdot (x^2)^{3-2}\cdot (-y^2)^2+^{3}C_3\cdot (x^2)^0\cdot (-y^2)3$

$(x^2+(-y^2))^3=^{3}C_0\cdot (x^2)^{3}\cdot (-y^2)^0+^{3}C_1\cdot (x^2)^{2}\cdot (-y^2)^1+^{3}C_2\cdot (x^2)^{1}\cdot (-y^2)^2+^{3}C_3\cdot (x^2)^0\cdot (-y^2)3$

$(x^2+(-y^2))^3=\frac{3!}{3!\cdot 0!}\cdot x^6+\frac{3!}{2!\cdot 1!}\cdot x^4\cdot (-y^2)+\frac{3!}{2!\cdot 1!}\cdot x^2\cdot y^4+\frac{3!}{0!\cdot 3!}\cdot (-y^6)$

$(x^2+(-y^2))^3=x^6+3\cdot x^4\cdot (-y^2)+3\cdot x^2\cdot y^4+3\cdot (-y^6)$

4:

We need to find the fourth term we have a formula:

$T_{r+1}=\sum^{n}C_r\cdot a^{n-r}\cdot b^r$

Here, $a=a,b=-b^2,r=3\text{and}n=5$  on substituting the values we get:

$T_{3+1}=\sum^{5}C_3\cdot a^{5-3}\cdot (-b^2)^3$

Using $^{n}C_r=\frac{n!}{r!\cdot (n-r)!}$ we get:

$T_{3+1}=\frac{5!}{3!\cdot 2!}\cdot a^{2}\cdot -b^6$

$T_{3+1}=10\cdot a^{2}\cdot -b^6$  is the fourth term.