Solution (1):
We use the substitution method here.
Let a=2000 and b=11.
Thus the expression becomes:
^{3}-a^{3}-b^{3}}{(a+b)ab}=\frac{3a^{2}b+3ab^{2}}{(a+b)ab}=\frac{3ab(a+b)}{(a+b)ab}= 3 "\frac{(a+b)^{3}-a^{3}-b^{3}}{(a+b)ab}=\frac{3a^{2}b+3ab^{2}}{(a+b)ab}=\frac{3ab(a+b)}{(a+b)ab}= 3")
Solution(2):
Draw a right angled triangle and label one of the acute angles x degrees.
We use the substitution method here.
Let a=2000 and b=11.
Thus the expression becomes:
Solution(2):
Draw a right angled triangle and label one of the acute angles x degrees.
Cos(x) refers to adjacent side/hypotenuse
Sin(x) refers to opposite side/hypotenuse
By Pythagoras' theorem,
we have
therefore