Question

find the following product:

Answer 1

Hope you find your answer.

Answer 2

Answer: The calculations are done below.

Step-by-step explanation: We will be using the following formula in the given  product simplifications:

$(a-b)(a+b)=a^2-b^2.$

The simplifications are as follows:

(1) We have

(i) $(x+2)(x-2)=x^2-2^2=x^2-4.$

(ii) $(d-8)(d+8)=d^2-8^2=d^2-64.$

(iii) $(5+9m)(5-9m)=5^2-(9m)^2=25-81m^2.$

(iv) $(x^2-y^2)(x^2+y^2)=(x^2)^2-(y^2)^2=x^4-y^4.$

(v) $\left(\dfrac{2}{5}ab-c\right)\left(\dfrac{2}{5}ab+c\right)=\left(\dfrac{2}{5}ab\right)^2-c^2=\dfrac{4}{25}ab-c^2.$

(vi) $\left(\dfrac{2}{b}-\dfrac{5}{c}\right) \left(\dfrac{2}{b}+\dfrac{5}{c}\right)=\left(\dfrac{2}{b}\right)^2-\left(\dfrac{5}{c}\right)^2=\dfrac{4}{b^2}-\dfrac{25}{c^2}.$

(2) We have

$(i)~(a+1)(a-1)(a^2+1)\\\\=(a^2-1^2)(a^2+1)\\\\=(a^2-1)(a^2+1)\\\\=(a^2)^2-1^2\\\\=a^4-1.$

$(ii)~(a+x)(a-x)(a^2+x^2)\\\\=(a^2-x^2)(a^2+x^2)\\\\=(a^2)^2-(x^2)^2\\\\=a^4-x^4.$

$(iii)~\left(x-\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}\right)\\\\\\=\left(x^2-\left(\dfrac{1}{x}\right)^2\right)\left(x^2+\dfrac{1}{x^2}\right)\\\\\\=\left(x^2-\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}\right)\\\\\\=(x^2)^2-\left(\dfrac{1}{x^2}\right)^2\\\\\\=x^4-\dfrac{1}{x^4}.$

$(iv)~\left(x^3-\dfrac{1}{x^3}\right)\left(x^3+\dfrac{1}{x^3}\right)\left(x^6+\dfrac{1}{x^6}\right)\\\\\\=\left((x^3)^2-\left(\dfrac{1}{x^3}\right)^2\right)\left(x^6+\dfrac{1}{x^6}\right)\\\\\\=\left(x^6-\dfrac{1}{x^6}\right)\left(x^6+\dfrac{1}{x^6}\right)\\\\\\=(x^6)^2-\left(\dfrac{1}{x^6}\right)^2\\\\\\=x^{12}-\dfrac{1}{x^{12}}.$

Thus, all the products are simplified.