question_answer

(a) Add: \[p\text{ }(p\text{ }-\text{ }q),\text{ }q\text{ }\left( q\text{ }-\text{ }r \right)\]and \[r\text{ }\left( r\text{ }-\text{ }p \right)\]

(b) Add: \[2x\left( z-x-y \right)\] and \[2y\left( z-y-x \right)\]

(c) Subtract: 31 \[(I-4m+5n)\]from \[4l\left( 10n-3m+2l \right)\]

(d) Subtract: \[3a\left( a+\text{b}+c \right)-2b\text{ }\left( a-b+c \right)\] from \[4c\text{ }\left( -a\text{ }+\text{ b }+\text{ }c \right)\]

Answer:

(a)   First expression \[=p\left( p-q \right)={{p}^{2}}-\text{ }pq\]

Second expression \[=q(q-r)={{q}^{2}}-qr\]

Third expression \[=r(r-p)={{r}^{2}}-rp\]

Now, adding three expression

\[\begin{align}   & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{p}^{2}}-pq \\  & +\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+{{q}^{2}}-qr \\  & +\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+{{r}^{2}}-rp \\  & \overline{\,\underline{\,\,\,\,\,\,\,\,\,{{p}^{2}}-pq+{{q}^{2}}-qr+{{r}^{2}}-rp\,\,\,\,\,\,\,\,\,\,\,}\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \end{align}\]

Therefore, \[{{p}^{2}}+\text{ }{{\text{q}}^{2}}+\text{ }{{r}^{2}}-\text{ }\left( pq+qr+rp \right)\]

(b) First expression

\[~=\text{ }2x\left( \text{z}-x-y \right)=2xz-2{{x}^{2}}-2xy\]

Second expression

\[=\text{ }2y\left( z-y-x \right)=2yz-2{{y}^{2}}-2xy\]

Now, adding the two expressions

\[\begin{align}   & \,\,\,\,\,\,\,\,2xz-2{{x}^{2}}-2xy \\  & \underline{+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-2xy+2yz-2{{y}^{2}}}\,\, \\  & \underline{2xz-2{{x}^{2}}-4xy+2yz-2{{y}^{2}}} \\ \end{align}\]

Therefore, \[-\text{ }2{{x}^{2}}-2{{y}^{2}}-4xy+2xz+2yz\]

(c) We have,          \[31(I-4m+5n)\]

and          \[l\left( 10n-3m+2l \right)=40ln-12lm+8{{l}^{2}}\]

Now,           \[\begin{align}   & 15\ln -\,\,\,12lm\,+\,3{{l}^{2}} \\  & (-)\,\,\,\,\,\,\,(+)\,\,\,\,\,\,(-) \\  & \underline{\overline{\,\,25\ln \,+\,\,0+\,\,\,5{{l}^{2}}}}\, \\ \end{align}\]

Therefore,                \[5{{l}^{2}}+25ln\]

(d) We have, \[3a\left( a+b+c \right)-2b\text{ }\left( a-b+c \right)\]

\[=3{{a}^{2}}+3ab+3ac-\left( 2ab-2{{b}^{2}}+2bc \right)\]

\[=3{{a}^{2}}+3ab+3ac-2ab+2{{b}^{2}}-2bc)\]

\[=3{{a}^{2}}+2{{b}^{2}}+ab+3ac-2bc\]

and   \[4c(-\text{ }a+b+c)=-\text{ }4ac+\text{ }4bc\text{ }+\text{ }4{{c}^{2}}\]

Now,

\[-4ac+4bc+4{{c}^{2}}-(3{{a}^{2}}+2{{b}^{2}}+ab+3ac-2bc)\]

\[=-\,4ac+4bc+4{{c}^{2}}-3{{a}^{2}}-2{{b}^{2}}-ab-3ac+2bc)\]

\[=-3{{a}^{2}}-2{{b}^{2}}+4{{c}^{2}}-ab-\text{7}ac+6bc\]

Therefore, \[-\text{ }3{{a}^{2}}-2{{b}^{2}}+\text{ }4{{c}^{2}}-\text{ }ab+6bc-7ac\]

\[=3{{l}^{2}}-12lm+15In\]