Sure, we can expand the expression using binomial theorem, which is really powerful when dealing with expressions like this.
Let's start with the binomial expression to the power of 5, i.e., (2x³−3y)⁵.
Using the Binomial Theorem, we can expand it as follows:
= (2x³)⁵ - 5 * (2x³)⁴ * (3y) + 10 * (2x³)³ * (3y)² - 10 * (2x³)² * (3y)³ + 5 * (2x³) * (3y)⁴ - (3y)⁵
which simplifies to get:
= 32x¹⁵ - 240x¹²y + 720x⁹y² - 1080x⁶y³ + 810x³y⁴ - 243y⁵
So in the expanded form of (2x³−3y)⁵,
the term with x⁶y³ is -1080x⁶y³
The coefficient of x⁶ in -1080x⁶y³ is -1080 and the coefficient of y³ is 1.
Now, we need to find the coefficient of the x⁶y³ term in the expansion. By coefficient, we mean the numerical or constant term that is associated with the variables.
So, from -1080x⁶y³, the coefficient of the x⁶y³ term is -1080 * 1 = -1080.
However, in this particular code example, you specifically request the coefficient of x²y. You'll notice that there is no x²y term in the expansion, so its coefficient is 0.
In conclusion, the coefficient of the x⁶y³ term is -1080 and the coefficient of x²y term is 0 in the expansion of (2x³−3y)⁵.
Answer: coefficient of x⁶y³ : -1080, coefficient of x²y : 0