Question

Show that: ${(4\mathrm{pq}+3\mathrm{q})}^2-{(4\mathrm{pq}-3\mathrm{q})}^2=48{\mathrm{pq}}^2$

Answer 1

To prove:${(4\mathrm{pq}+3\mathrm{q})}^2-{(4\mathrm{pq}-3\mathrm{q})}^2=48{\mathrm{pq}}^2$

$\mathrm{L}.\mathrm{H}.\mathrm{S}=$${(4\mathrm{pq}+3\mathrm{q})}^2-{(4\mathrm{pq}-3\mathrm{q})}^2$

$={{\left[\right(4\mathrm{pq})}^2+\left(3\mathrm{q}\right)}^2+2\times 4\mathrm{pq}\times 3\mathrm{q}]-{[{\left(4\mathrm{pq}\right)}^2+(3\mathrm{q})}^2-2\times 4\mathrm{pq}\times 3\mathrm{q}]$ [ Formula used: ${(\mathrm{a}+\mathrm{b})}^2={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}$ & ${(\mathrm{a}-\mathrm{b})}^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}$ ]

$={{\left(4\mathrm{pq}\right)}^2+\left(3\mathrm{q}\right)}^2+2\times 4\mathrm{pq}\times 3\mathrm{q}-{{\left(4\mathrm{pq}\right)}^2-\left(3\mathrm{q}\right)}^2+2\times 4\mathrm{pq}\times 3\mathrm{q}$

$=24{\mathrm{pq}}^2+24{\mathrm{pq}}^2$ [Like terms with opposite sign cancel each other]

$=48{\mathrm{pq}}^2$

⇒$\mathrm{L}.\mathrm{H}.\mathrm{S}=$$\mathrm{R}.\mathrm{H}.\mathrm{S}$

Thus,${(4\mathrm{pq}+3\mathrm{q})}^2-{(4\mathrm{pq}-3\mathrm{q})}^2=48{\mathrm{pq}}^2$