 Hi and welcome to the session. I am Asha and I am going to help you solve the following problem which says write the following cubes in expanded form. So first let us learn some simple identities with the help of which we will expand the given following four cubes. The first identity is u plus v whole cube is equal to u cube plus v cube plus 3 into u into v plus and the second identity is v whole cube is equal to u cube minus v cube minus 3 u v into u minus v and on opening the brackets the above identity can be written as u cube plus v cube plus 3 u square v plus 3 u v square and the second one can be written as u cube minus v cube minus 3 u square v plus 3 u v square. So these two identities are a key idea with the help of which we will expand the following given cubes. Let us now start with the solution. First we have to find the cube of 2h plus 1. Now I am comparing it with the first identity here. u is equal to 2x and v is equal to 1 and on applying it the second identity is u cube that is 2x whole cube plus v cube which is 1 cube plus 3 into u square v so u is 2x whole square into v is 1 plus 3 u v square. So u is 2x and v is 1. Now 2x whole cube is 8x cube 1 whole cube is 1 plus 3 into 2x whole square is 4x square into 1 plus 3 into 2 is 6 with x. Now for the simplifying we have 8x cube plus 1 plus 3 cos 12x square plus 6x and hence on expanding we get the answer as 8x cube plus 12x square plus 6x plus 1. So this completes the first part and now proceeding on to the second part which is 2a minus 3b whole cube. Now on comparing it with the left hand side of the second identity we find here that u is equal to 2a and v is equal to 3b and thus on applying the identity first we have u square that is 2a whole cube and we have minus v cube that is 3b whole cube minus 3 times of u v square 2a whole square into v that is 3b plus 3 u into v square that is 3b whole square. Now 2a whole cube is 8a cube minus 3b whole cube is 27b cube minus 3 2a whole square is 4a into 3b that is 4a square 2 into 2a is 2 2a 4 and a into a is a square plus 3 into 2a and 3b whole square is 9b square which is further equal to 8a cube minus 27b cube minus 4 3s are 12 and 12 3s are 36 a square b plus 3 2s are 6 and 6 lines are 54ab square and hence on expanding 2a minus 3b whole cube we get 8a cube minus 27b cube minus 36a square b plus 54ab square. So this is the answer of the second part now proceeding on to the third part where we have to expand 3 upon 2 into x plus 1 whole cube. Now comparing it with the first identity here we find that u is equal to 3 upon 2 x and v is equal to 1 and thus on applying identity 1 which is our key idea we have 3 upon 2 into x whole cube then we have v cube which is 1 cube plus 3 times of u square v that is 3 upon 2 into x whole square into v which is 1 plus 3 uv square so u is 3 upon 2 into x and v is 1 square. Now 3 whole cube is 27 2 whole cube is 8 and x whole cube is x cube then we have plus sign and 1 whole cube is 1 plus 3 3 upon 2 x whole square is 9 upon 4 x square into 1 plus 3 into 3 upon 2 into x which is further equal to 27 upon 8 x cube plus 1 plus 9 into 3 is 27 upon 4 x square plus 3 into 3 is 9 and we have x upon 2 and thus on expanding 3 upon 2 into x plus 1 whole cube we get an answer as 27 upon 8 x cube plus 1 plus 27 upon 4 x square plus 9 upon 2 x which can also be written as 27 upon 8 x cube plus 27 upon 4 x square plus 9 upon 2 x plus 1. So this completes the third part and now proceeding on to the last part where we have to expand x minus 2 upon 3 y whole cube. Now comparing it with the left hand side of the second identity we find here that u is equal to x and v is equal to 2 upon 3 y and on applying the identity this can be written as first u cube so x whole cube minus v cube v is 2 upon 3 y whole cube that we have minus 3 u square v so u is x square and v is 2 upon 3 into y then plus 3 u v square so u is x and v is 2 upon 3 into y whole square which is further equal to x cube 2 upon 3 y whole cube is 8 upon 27 y cube minus this 3 cancels out with 3 and we have 2 x square y plus 3 into x and 2 upon 3 y whole square is 4 upon 9 y square. Now 3 is the common factor of the numerator and denominator so on cancelling we have 3 1's are 3 and 3 3's are 9 thus this can further be written as x cube minus 8 upon 27 y cube minus 2 x square y plus 4 upon 3 x y square hence on expanding then x minus 2 upon 3 into y whole cube we get the answer as x cube minus 8 upon 27 y cube minus 2 x square y plus 4 upon 3 x y square so this completes the last part hence the session so hope you enjoyed it and remember the identities we have used in this problem by solving them. Have a good day.