The value of \(\frac{(1.01)^3+0.000001}{1.0201-(0.01)^2}\times \frac{(7.85)^2-4.6225}{7.85-2.15}\) is

Given:

a = 1.01, b = 0.01, x = 7.85, y = 2.15

Formula used:

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

\(x^2-y^2=(x+y)(x-y)\)

Calculations:

For 1st term : \(\frac{(1.01)^3+0.000001}{1.0201-(0.01)^2}\)

⇒ a + b = 1.01 + 0.01 = 1.02 

⇒ a − b = 1.01 − 0.01 = 1

⇒ a² = 1.0201, ab = 1.01 × 0.01 = 0.0101, b² = 0.0001

⇒ a² − ab + b² = 1.0201 − 0.0101 + 0.0001 = 1.0101

⇒ \(\frac{a^3+b^3}{a^2-b^2}=\frac{(a+b)(a^2-ab+b^2)}{(a+b)(a-b)}=\frac{a^2-ab+b^2}{a-b}=\frac{1.0101}{1}=1.0101\)

For 2nd term: \(\frac{(7.85)^2-4.6225}{7.85-2.15}\)

⇒ x + y = 7.85 + 2.15 = 10

⇒ x − y = 5.7

⇒ \(\frac{x^2-y^2}{x-y}=\frac{(x+y)(x-y)}{x-y}=x+y=10\)

⇒ Result=1.0101 × 10 = 10.101

∴ The value of the expression is 10.101.