Write the following cubes in expanded form : $(2 a-3 b)^{3}$
Write the following cubes in expanded form : $(2 a-3 b)^{3}$
Using Identity $VI$ and Identity $VII,$ we have
$(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y),$ and $(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)$
$(2 a-3 b)^{3}=(2 a)^{3}-(3 b)^{3}-3(2 a)(3 b)[(2 a)-(3 b)]$
$=8 a ^{3}-27 b ^{3}-18 ab (2 a -3 b )$ [Using Identity $VII$]
$=8 a^{3}-27 b^{3}-\left[36 a^{2} b-54 a b^{2}\right]=8 a^{3}-27 b^{3}-36 a^{2} b+54 a b^{2}$
Similar Questions
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given : $\boxed{\rm {Area}\,:35{y^2}+ 13y - 12}$
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given : $\boxed{\rm {Area}\,:35{y^2}+ 13y - 12}$
Factorise : $12 x^{2}-7 x+1$
Factorise : $12 x^{2}-7 x+1$
Evaluate using suitable identities : $(104)^{3}$
Evaluate using suitable identities : $(104)^{3}$
Find the zero of the polynomial : $p(x)=a x,\,\, a \neq 0$
Find the zero of the polynomial : $p(x)=a x,\,\, a \neq 0$
Factorise : $8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}$
Factorise : $8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}$