Question

Simplify:
(i) $(a^2-b^2{)}^2$
(ii) $(2x+5{)}^2-(2x-5{)}^2$
(iii) $(7m-8n{)}^2+(7m+8n{)}^2$
(iv) $(4m+5n{)}^2+(4n+5m{)}^2$

(v) $(2.5p-1.5q{)}^2-(1.5p-2.5q{)}^2$
(vi) $(ab+bc{)}^2-2a{b}^{2}c$
(vii) $(m^2-n^{2}m{)}^2+2m^3{n}^2$

Answer 1

i) $(a^2-b^2{)}^2=(a^2{)}^2+(b^2{)}^2-2(a^2\left)\right(b^2)$ $\because \left[\right(a+b{)}^2$$=a^2+b^2+2ab]$

$=a^4+b^4-2a^2{b}^2$

ii) $(2x+5{)}^2-(2x-5{)}^2$

Let $A=2x+5$ and $B=2x-5$

$\therefore (2x+5{)}^2-(2x-5{)}^2=A^2-B^2$

$=(A+B)(A-B)$ $[\because a^2-b^2=(a+b\left)\right(a-b\left)\right]$

$=[2x+5+(2x-5\left)\right][2x+5-(2x-5\left)\right]$

$=[2x+5+2x-5][2x+5-2x+5]$

$=\left(4x\right)\left(10\right)$

$=40x$

iii) $(7m-8n{)}^2+(7m+8n{)}^2$

$=(7m{)}^2+(8n{)}^2-2\left(7m\right)\left(8n\right)+(7m{)}^2+(8n{)}^2+2\left(7m\right)\left(8n\right)$

$=49m^2+64n^2-112mn+49m^2+64n^2+112mn$

$=98m^2+128n^2$

iv) $(4m+5n{)}^2+(5m+4n{)}^2$

$=(4m{)}^2+(5n{)}^2+2\left(4m\right)\left(5n\right)+(5m{)}^2+(4n{)}^2+2\left(5m\right)\left(4n\right)$ $[\because (a-b{)}^2=a^2-2ab+b^2]$

$=16m^2+25n^2+40mn+25m^2+16n^2+40mn$

$=41m^2+41n^2+80mn$

v) $(2.5p-1.5q{)}^2-(1.5p-2.5q{)}^2$

$=(2.5p{)}^2+(1.5q{)}^2-2\left(2.5p\right)\left(1.5q\right)-\left[\right(1.5p{)}^2+(2.5q{)}^2-2(1.5p\left)\right(2.5q\left)\right]$

$=6.25p^2+2.25q^2-7.5pq-2.25p^2-6.25q^2+7.5pq$

$=4p^2-4q^2$

$=4(p^2-q^2)$

$=4(p-q)(p+q)$

vi) $(ab+bc{)}^2-2a{b}^{2}c$

$=(ab{)}^2+(bc{)}^2+2\left(ab\right)\left(bc\right)-2a{b}^{2}c$

$=a^2{b}^2+b^2{c}^2+2a{b}^{2}c-2a{b}^{2}c$

$=b^2(a^2+c^2)$

vii) $(m^2-n^{2}m{)}^2+2m^3{n}^2$

$=(m^2{)}^2+(n^{2}m{)}^2-2\left(m^2\right)\left(n^{2}m\right)+2m^3{n}^2$

$=m^4+m^2{n}^4-2m^3{n}^2+2m^3{n}^2$

$=m^2(m^2-n^4)$