Use $(a+b)^2=a^2+2ab+b^2$ in each part. For example, in (ii), $a=\dfrac{7}{5}x$ and $b=\dfrac{3}{2}y$, so the middle term is $2ab=2\cdot\dfrac{7}{5}x\cdot\dfrac{3}{2}y=\dfrac{21}{5}xy$. The other parts follow similarly by squaring the two terms and adding twice their product.
(i) $49x^2 + 56xy + 16y^2$
(ii) $\dfrac{49}{25}x^2 + \dfrac{21}{5}xy + \dfrac{9}{4}y^2$
(iii) $6.25p^2 + 7.5pq + 2.25q^2$
(iv) $\dfrac{9}{16}s^2 + 12st + 64t^2$
(v) $x^2 + \dfrac{x}{y} + \dfrac{1}{4y^2}$
(vi) $\dfrac{1}{x^2} + \dfrac{2}{xy} + \dfrac{1}{y^2}$
$(64)^2=(60+4)^2=3600+480+16=4096$. $(105)^2=(100+5)^2=10000+1000+25=11025$. $(205)^2=(200+5)^2=40000+2000+25=42025$.
(i) $4096$
(ii) $11025$
(iii) $42025$
Compare each expression with $a^2+2ab+b^2$. In (v), take out $\dfrac{1}{3}$ to get $\dfrac{1}{3}(9a^2+12ab+4b^2)=\dfrac{1}{3}(3a+2b)^2$. In (vi), take out $\dfrac{1}{5}$ to get $\dfrac{1}{5}(9s^2+30sv+25v^2)=\dfrac{1}{5}(3s+5v)^2$.
(i) $(3x+4y)^2$
(ii) $(2s+5t)^2$
(iii) $(7x+2y)^2$
(iv) $\left(8p+\dfrac{2}{3}q\right)^2$
(v) $\dfrac{1}{3}(3a+2b)^2$
(vi) $\dfrac{1}{5}(3s+5v)^2$
$(79)^2=(80-1)^2=6400-160+1=6241$. $(193)^2=(200-7)^2=40000-2800+49=37249$. $(299)^2=(300-1)^2=90000-600+1=89401$.
(i) $6241$
(ii) $37249$
(iii) $89401$
Use nearby round numbers: $117^2=(100+17)^2=13689$; $78^2=(80-2)^2=6084$; $198^2=(200-2)^2=39204$; $214^2=(200+14)^2=45796$; $1104^2=(1100+4)^2=1218816$; $1120^2=(1100+20)^2=1254400$.
(i) $13689$
(ii) $6084$
(iii) $39204$
(iv) $45796$
(v) $1218816$
(vi) $1254400$
Each expression matches either $(a-b)^2$ or $(a+b+c)^2$. For (iii), the square terms are $\left(\dfrac{m}{3}\right)^2$, $\left(\dfrac{k}{2}\right)^2$ and $(3n)^2$, and the cross terms are $\dfrac{mk}{3}$, $3nk$ and $2mn$. For (v), $(3a-2b+c)^2$ gives $9a^2+4b^2+c^2-12ab+6ac-4bc$.
(i) $(4y-3)^2$
(ii) $\left(\dfrac{3}{2}s+2t\right)^2$
(iii) $\left(\dfrac{m}{3}+\dfrac{k}{2}+3n\right)^2$
(iv) $\left(\dfrac{p}{4}-\dfrac{4}{p}\right)^2$
(v) $(3a-2b+c)^2$
For (i), square $p,3q,7r$ and add twice each pairwise product. For (ii), take the three terms as $3x,-2y,4z$; the cross terms are $2(3x)(-2y)=-12xy$, $2(-2y)(4z)=-16yz$ and $2(3x)(4z)=24xz$.
(i) $p^2+9q^2+49r^2+6pq+42qr+14pr$
(ii) $9x^2+4y^2+16z^2-12xy-16yz+24xz$
A single counterexample is enough. Put $a=b=c=1$. Then the left side is $(1+1-1)^2+(1-1+1)^2+(1-1-1)^2=1+1+1=3$, while the right side is $2+2+2=6$. Since the two sides are not equal for these values, the statement is not an identity.
No, it is not an identity.
Split the middle term or use factor pairs. For example, $s^2-11s+24$ needs two numbers with sum $-11$ and product $24$, namely $-3$ and $-8$. Similarly, $3x^2-4x-7=(3x-7)(x+1)$, $10x^2-11x-6=(2x-3)(5x+2)$ and $6x^2+7x+2=(3x+2)(2x+1)$.
(i) $(s-3)(s-8)$
(ii) $(3x-7)(x+1)$
(iii) $(2x-3)(5x+2)$
(iv) $(3x+2)(2x+1)$
Use nearby convenient numbers: $41^2=(40+1)^2=1681$; $27^2=(30-3)^2=729$; $23\times17=(20+3)(20-3)=400-9=391$; $135^2=(100+35)^2=18225$; $97^2=(100-3)^2=9409$; $18\times29=522$; $34\times43=1462$; $205^2=(200+5)^2=42025$.
(i) $1681$
(ii) $729$
(iii) $391$
(iv) $18225$
(v) $9409$
(vi) $522$
(vii) $1462$
(viii) $42025$
Parts (i), (ii), (iv) and (v) are perfect squares. For example, $(8u-11v-2w)^2=64u^2+121v^2+4w^2-176uv-32uw+44vw$. For (iii), the two numbers with product $-42$ and sum $-1$ are $-7$ and $6$, so $r^2-r-42=(r-7)(r+6)$.
(i) $(3a-b+2c)^2$
(ii) $(4s-5t)^2$
(iii) $(r-7)(r+6)$
(iv) $(7g+h)^2$
(v) $(8u-11v-2w)^2$
Factor and cancel common non-zero factors. In (ii), the numerator is $(n-m)^3$ and the denominator is $5(n-m)^2$, so the result is $(n-m)/5$. In (iv), the numerator is $(2y-5z)^2$ and the denominator is $(5z-2y)(5z+2y)$, giving $\dfrac{5z-2y}{2y+5z}$. In (v), all four quadratic factors cancel. In (vi), $p^4-16=(p^2-4)(p^2+4)=(p-2)(p+2)(p^2+4)$ and the denominator is $(p-2)^2$.
(i) $\dfrac{3(p-3q)(p+2q)}{(p+5q)(p-2q)}$
(ii) $\dfrac{n-m}{5}$
(iii) $\dfrac{w^2+v^2+x^2+wv-vx-wx}{w-v+x}$
(iv) $\dfrac{5z-2y}{2y+5z}$
(v) $1$
(vi) $\dfrac{(p+2)(p^2+4)}{p-2}$
Use $(a-b)^2$, $(a+b)(a-b)=a^2-b^2$, $a^3-b^3=(a-b)(a^2+ab+b^2)$ and $(a-b)^3=a^3-3a^2b+3ab^2-b^3$ as needed. For example, part (v) is $s^3-(2t)^3=s^3-8t^3$, and part (ix) uses $a=\dfrac{7}{2}k$, $b=\dfrac{2}{3}m$ in $(a-b)^3$.
(i) $9x^2-24x+16$
(ii) $4s^2-49$
(iii) $p^4-\dfrac{1}{4}$
(iv) $4n^2-49$
(v) $s^3-8t^3$
(vi) $\dfrac{1}{4r^2}-4+16r^2$
(vii) $9m^2+16k^2+l^2-24mk+6ml-8kl$
(viii) $x^3-x^2y+\dfrac{1}{3}xy^2-\dfrac{1}{27}y^3$
(ix) $\dfrac{343}{8}k^3-\dfrac{49}{2}k^2m+\dfrac{14}{3}km^2-\dfrac{8}{27}m^3$
For products near a common centre, use $(a+b)(a-b)=a^2-b^2$: $17\times21=(19-2)(19+2)=361-4=357$, $104\times96=(100+4)(100-4)=10000-16=9984$, and $24\times16=(20+4)(20-4)=400-16=384$. For cubes, use $(a\pm b)^3$ with nearby round numbers, e.g. $147^3=(150-3)^3=3176523$.
(i) $357$
(ii) $9984$
(iii) $384$
(iv) $3176523$
(v) $7880599$
(vi) $2048383$
(vii) $-1225043$
(viii) $-26730899$
Match each expression to a standard identity. (i) $(a+b)^2$ with $a=2y,\ b=\dfrac{1}{4y}$. (ii) $a^2-b^2=(a-b)(a+b)$. (iii) $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=3b,\ b=\dfrac{1}{4b}$. (iv) Split the middle term: $x^2+\dfrac{5x}{6}+\dfrac{1}{6}=\left(x+\dfrac12\right)\left(x+\dfrac13\right)$. (v) $(a-b)^3=a^3-3a^2b+3ab^2-b^3$ with $a=3u,\ b=\dfrac15$ gives $27u^3-\dfrac{27u^2}{5}+\dfrac{9u}{25}-\dfrac{1}{125}=\left(3u-\dfrac15\right)^3$. (vi) $a^3+b^3=(a+b)(a^2-ab+b^2)$ with $a=4y,\ b=\dfrac{z}{5}$. (vii) $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ with $a=p,\ b=3q,\ c=r$. (viii) $(3m-2)^2$. (ix) Take out $\dfrac13$ and apply the same $a^3+b^3+c^3-3abc$ identity. (x) For $(a+b+c)^2$ the cross terms must be $2ab,2bc,2ca$; with $a=2x,b=3y,c=6z$ that is $12xy,36yz,24xz$, so the printed $24xy$ and $12xz$ are interchanged — the intended square is $(2x+3y+6z)^2$. (xi) $(a-b)^3$ with $a=3u,\ b=\dfrac16$ gives $\left(3u-\dfrac16\right)^3$.
(i) $\left(2y+\dfrac{1}{4y}\right)^2$
(ii) $\left(3m-\dfrac{1}{5n}\right)\left(3m+\dfrac{1}{5n}\right)$
(iii) $\left(3b-\dfrac{1}{4b}\right)\left(9b^2+\dfrac{3}{4}+\dfrac{1}{16b^2}\right)$
(iv) $\left(x+\dfrac{1}{2}\right)\left(x+\dfrac{1}{3}\right)$
(v) $\left(3u-\dfrac{1}{5}\right)^3$
(vi) $\left(4y+\dfrac{z}{5}\right)\left(16y^2-\dfrac{4}{5}yz+\dfrac{z^2}{25}\right)$
(vii) $(p+3q+r)(p^2+9q^2+r^2-3pq-3qr-pr)$
(viii) $(3m-2)^2$
(ix) $\dfrac{1}{3}(3x-2y+z)(9x^2+4y^2+z^2+6xy+2yz-3xz)$
(x) As printed, the $xy$ and $xz$ coefficients are interchanged from a perfect square. The intended expression $4x^2+9y^2+36z^2+12xy+36yz+24xz$ factors as $(2x+3y+6z)^2$.
(xi) $\left(3u-\dfrac{1}{6}\right)^3$
(i) $4x^2+4x+1=(2x+1)^2$ and $4x^2-1=(2x-1)(2x+1)$. (ii) $3a^3-24b^3=3(a^3-8b^3)=3(a-2b)(a^2+2ab+4b^2)$ and the denominator is $9(a-2b)(a+2b)$. (iii) $s^3+125t^3=(s+5t)(s^2-5st+25t^2)$ and $s^2-2st-35t^2=(s+5t)(s-7t)$.
(i) $\dfrac{2x+1}{2x-1}$
(ii) $\dfrac{3(a^2+2ab+4b^2)}{a+2b}$
(iii) $\dfrac{s^2-5st+25t^2}{s-7t}$
Factor the areas. $25a^2-30ab+9b^2=(5a-3b)^2$, so one possible rectangle is a square with side $5a-3b$. Also $36s^2-49t^2=(6s)^2-(7t)^2=(6s+7t)(6s-7t)$.
(i) Length $=5a-3b$, breadth $=5a-3b$.
(ii) Length $=6s+7t$, breadth $=6s-7t$.
Factor each volume into three factors. $6a^2-24b^2=6(a^2-4b^2)=6(a-2b)(a+2b)$. Also $3ps^2-15ps+12p=3p(s^2-5s+4)=3p(s-1)(s-4)$.
(i) Possible dimensions are $6$, $(a-2b)$ and $(a+2b)$.
(ii) Possible dimensions are $3p$, $(s-1)$ and $(s-4)$.