Question

Expand: (i) $(7x+8y{)}^3$ (ii) $101{)}^3$

Answer 1

(i) Let us simplify the epxression, we get
$(7x+8y{)}^3=(7x{)}^3+[3\times (7x{)}^2\times \left(8y\right)]+[3\times \left(7x\right)\times \left(8y{)}^2\right]+(8y{)}^3$
By using the formula,
$(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3$
Here $a=7x,b=8y$
Now, substituting the above values, we get
$=343x^3+(3\times 49\times 8)x^{2}y+(3\times 7\times 64)x{y}^2+512y^3$
$=343x^3+1176x^{2}y+1344x{y}^2+512y^3$
$\therefore (7x+8y{)}^3=343x^3+1176x^{2}y+1344x{y}^2+512y^3$

(ii) Let us simplify the expression, we get
$(101{)}^3=(100+1{)}^3$
$(100+1{)}^3=(100{)}^3+[3\times (100{)}^2\times \left(1\right)]+[3\times \left(100\right)\times \left(1{)}^2\right]+(1{)}^3$
By using the formula,
$(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3$
Here $a=100,b=1$
Now, substituting the above values, we get
$=1000000+(3\times 10000\times 1)+(3\times 100\times 1)+1$
$=1000000+30000+300+1$
$=1030301$
$\therefore (101{)}^3=(100+1{)}^3=1030301$