 Hello and welcome to the session. In this session we discussed the following question which says, by using the properties of determinants, show that determinant with elements a square bc, ac plus c square, a square plus ab, b square ac, ab, b square plus bc, c square is equal to 4 a square b square c square. Now we move on to the solution. We need to show that determinant with elements a square bc, ac plus c square, a square plus ab, b square, ac, ab, b square plus bc, c square is equal to 4 a square b square c square. We consider the LHS. We take delta equal to LHS that is the determinant with elements a square bc, ac plus c square, a square plus ab, b square, ac, ab, b square plus bc, c square. Now taking common factors abc from c1, c2, c3 respectively we get delta is equal to abc into determinant with elements a, c, a plus c, a plus b, b, a, b, b plus c, c. Now applying c1 goes to c1 plus c2 minus c3 we get delta is equal to abc into determinant with elements 0, c, a plus c, 2b, b, a, 2b, b plus c, c. That is the elements of the column c2 and c3 remain the same and the elements of column c1 are given by c1 plus c2 minus c3. Now taking out common factor 2b from c1 we get delta is equal to abc into 2b into determinant with elements 0, c, a plus c, 1, b, a, 1, b plus c, c. Now we apply r2 goes to r2 minus r3 and on doing this we get delta is equal to 2ab square c into determinant with elements 0, c, a plus c, 0, minus c, a minus c, 1, b plus c, c. That is the elements of r1 and r3 remain same and the elements of r2 are given by r2 minus r3. Now expanding along c1 we get delta is equal to 2ab square c into, now the first element of c1 is 0, so 0 multiplied by anything would be 0, so 0 minus the second element of c1 which is 0 again plus the third element of c1 that is 1 into c multiplied by a minus c minus c multiplied by a plus c. And so we get delta is equal to 2ab square c into c into a minus c plus c into a plus c. Thus delta is equal to 2ab square c square we take c common, so this into a minus c plus a plus c that is delta is equal to 2ab square c square into a minus c plus a plus c. Now minus c and c chances, so we have delta is equal to 2ab square c square into 2a, this gives us delta is equal to 4a square b square c square which is equal to the rhs. Thus we get the rhs is equal to the rhs hence proved, so this completes the session hope you have understood the solution for this question.