 Hello and welcome to this session. Let us understand the following problem today. Evaluate the term in x, y, x plus y, y, x plus y, x, x, y, x, y. Now let us write this equation. We have the term in delta is equal to x, y, x plus y, y, x plus y, x, x plus y, x, y. Now, applying r1 tends to r1 plus r2 plus r3, we get here 2x plus 2y, 2x plus 2y, 2x plus 2y, rest r2 and r3 as it is. Now taking common 2 into x plus y from r1, therefore we get twice of x plus y common. So we are left with the determinant 1, 1, 1, y, x plus y, x, x plus y, x, y. Now applying c2 tends to c2 minus c1 and c3 tends to c3 minus c1. So we get twice of x plus y multiplied by determinant 1, column 1 as it is. So we write here y, x plus y. Now applying c2 minus c1, so we get here 0, x and minus y. Now applying c3 minus c1, so we get here 0, x minus y, minus x. Now it is equal to twice of x plus y. Now expanding through this one, eliminating this and this, we get multiplied by into minus x, minus x minus y into minus y. So which is equal to twice of x plus y into minus x squared plus y into x minus y, which is equal to twice of x plus y into minus x squared plus yx minus y squared. Which is equal to taking minus common from this, we get minus 2 into x plus y into x squared minus yx plus y squared, which is equal to minus 2 into x cubed plus y cubed because this is the formula for x cubed plus y cubed. So the required answer is twice of x minus x cubed plus y cubed, that is minus 2 into x cubed plus y cubed. This is our required answer. I hope you understood the problem. Bye and have a nice day.