anemone said:The numbers $a,\,b$ and $\dfrac{a^2+b^2+6}{ab}$ are positive integers. Prove that $\dfrac{a^2+b^2+6}{ab}$ is a perfect cube.
Albert said:let k=$\dfrac{a^2+b^2+6}{ab}---(1)$
then: $2+\dfrac{6}{ab}\leq k\leq \dfrac {a^2+b^2+6ab}{ab}=\dfrac {b}{a}+\dfrac {a}{b}+6$
for each $a,b\in N , max (2+\dfrac {6}{ab})=8,(a=b=1)$
and $min(\dfrac {b}{a}+\dfrac{a}{b}+6)=8 (for \,\, each \,\, a=b>0)$
if $a,b,k \in N, \,\,then\,\, k=8$
$\therefore k=8 $ is a perfect cube
To prove that (a²+b²+6)/(ab) is a perfect cube, we can use the method of contradiction. Assume that it is not a perfect cube and then simplify the expression to show that it leads to a contradiction. This will prove that the original expression is indeed a perfect cube.
Yes, for example, if we let a = 2 and b = 3, the expression becomes (2²+3²+6)/(2*3) = (4+9+6)/6 = 19/6. This is not a perfect cube since the cube root of 19 is not a whole number. However, if we simplify the expression to (a²+b²+6)/(ab) = (2+3) = 5, we can see that it is indeed a perfect cube since the cube root of 5 is equal to 5.
Proving that (a²+b²+6)/(ab) is a perfect cube allows us to understand the relationship between the variables a and b in the expression. It also allows us to make predictions and solve problems involving similar expressions.
The key steps in proving that (a²+b²+6)/(ab) is a perfect cube include simplifying the expression, assuming that it is not a perfect cube, and then arriving at a contradiction. This can be done by expanding the expression and using properties of perfect cubes to simplify it further.
In mathematics, a perfect cube is a number that can be expressed as the cube of an integer. This means that the number is the result of multiplying an integer by itself three times. For example, 8 is a perfect cube since it can be expressed as 2³, where 2 is the integer. In general, the nth perfect cube is represented by n³.