For the polynomial x⁶-64, we can apply the Difference of Squares but not the Difference of Cubes.
The Difference of Squares is a factoring pattern that can be used when we have the difference of two perfect squares, which means two terms that are both perfect squares and are being subtracted. In this case, x⁶-64 can be written as (x³)² - 8². This can be factored further as (x³ - 8)(x³ + 8).
However, the Difference of Cubes is a factoring pattern that can be used when we have the difference of two perfect cubes, which means two terms that are both perfect cubes and are being subtracted. Since x⁶-64 does not fit this pattern, we cannot apply the Difference of Cubes.
In summary, for the polynomial x⁶-64, we can apply the Difference of Squares by factoring it as (x³ - 8)(x³ + 8), but we cannot apply the Difference of Cubes.
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verify that the given differential equation is exact; then solve it. (9x^3 8y/x)dx (y^2 8lnx)dy=0
Given differential equation is:(9x^3 8y/x)dx (y^2 8lnx)dy=0.
If a differential equation is of the form M(x,y)dx + N(x,y)dy = 0, then it is called an exact differential equation
if:∂M/∂y = ∂N/∂x
Here, M = 9x³ + 8y/x and N = y² + 8lnx.
Therefore, ∂M/∂y = 8 and ∂N/∂x = 8/x.
Thus, the given differential equation is an exact differential equation.
Now, to find the solution of an exact differential equation, we integrate either M or N with respect to x or y, respectively.
Let's integrate M w.r.t x. So, we get:
∫Mdx = ∫(9x³ + 8y/x)dx= 9/4 x⁴ + 8y ln x + h(y) (put h(y) = 0,
since ∂(∂M/∂y)/∂y = ∂(∂N/∂x)/∂x )
Differentiating the above w.r.t y, we get:(d/dy) ∫Mdx = 8x + h'(y)
Comparing the above with N = y² + 8lnx
We get, h'(y) = y²∴ h(y) = y³/3 + c Here, c is a constant of integration.
The general solution of is 9/4 x⁴ + 8y ln x + y³/3 = c.
Yes the differential equation is exact
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