\displaystyle \textbf{Question 1: }\text{Expand:}
\displaystyle \text{(i) }(2a+3b+4c)^2\qquad \text{(ii) }(a+2b-5c)^2
\displaystyle \text{(iii) }(3x+2y-z)^2\qquad \text{(iv) }(3x-2y-1)^2
\displaystyle \text{(v) }\left(\frac{a}{2}-\frac{b}{3} +\frac{c}{4}\right)^2\qquad \text{(vi) }\left(x-\frac{2}{x}+3\right)^2
\displaystyle \text{Answer:}
\displaystyle \text{(i) }(2a+3b+4c)^2
\displaystyle =4a^2+9b^2+16c^2 +2\{(2a)(3b)+(3b)(4c)+(4c)(2a)\}
\displaystyle =4a^2+9b^2+16c^2+12ab+24bc+16ac.

\displaystyle \text{(ii) }(a+2b-5c)^2
\displaystyle =a^2+4b^2+25c^2 +2\{(a)(2b)+(2b)(-5c)+(-5c)(a)\}
\displaystyle =a^2+4b^2+25c^2+4ab-20bc-10ac.

\displaystyle \text{(iii) }(3x+2y-z)^2
\displaystyle =9x^2+4y^2+z^2 +2\{(3x)(2y)+(2y)(-z)+(-z)(3x)\}
\displaystyle =9x^2+4y^2+z^2+12xy-4yz-6xz.

\displaystyle \text{(iv) }(3x-2y-1)^2
\displaystyle =9x^2+4y^2+1 +2\{(3x)(-2y)+(-2y)(-1)+(-1)(3x)\}
\displaystyle =9x^2+4y^2+1-12xy+4y-6x.

\displaystyle \text{(v) }\left(\frac{a}{2}-\frac{b}{3} +\frac{c}{4}\right)^2
\displaystyle =\frac{a^2}{4}+\frac{b^2}{9}+\frac{c^2}{16} +2\left\{-\frac{ab}{6}-\frac{bc}{12}+\frac{ac}{8}\right\}
\displaystyle =\frac{a^2}{4}+\frac{b^2}{9}+\frac{c^2}{16} -\frac{ab}{3}-\frac{bc}{6}+\frac{ac}{4}.

\displaystyle \text{(vi) }\left(x-\frac{2}{x}+3\right)^2
\displaystyle =x^2+\frac{4}{x^2}+9 +2\left\{x\left(-\frac{2}{x}\right) +\left(-\frac{2}{x}\right)(3)+(3)(x)\right\}
\displaystyle =x^2+\frac{4}{x^2}+9 +2\left(-2-\frac{6}{x}+3x\right)
\displaystyle =x^2+\frac{4}{x^2}+9-4-\frac{12}{x}+6x
\displaystyle =x^2+\frac{4}{x^2}-\frac{12}{x}+6x+5.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }a+b+c=10\text{ and } ab+bc+ca=31,\text{ find }a^2+b^2+c^2.
\displaystyle \text{Answer:}
\displaystyle (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca).
\displaystyle \therefore a^2+b^2+c^2 =(a+b+c)^2-2(ab+bc+ca)
\displaystyle =10^2-2(31)=100-62=38.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }a+b+c=9\text{ and } a^2+b^2+c^2=29,\text{ find }ab+bc+ca.
\displaystyle \text{Answer:}
\displaystyle (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca).
\displaystyle \therefore 2(ab+bc+ca) =(a+b+c)^2-(a^2+b^2+c^2)
\displaystyle =9^2-29=81-29=52.
\displaystyle \therefore ab+bc+ca=26.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }a^2+b^2+c^2=45 \text{ and }ab+bc+ca=38,\text{ find }a+b+c.
\displaystyle \text{Answer:}
\displaystyle (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca).
\displaystyle \therefore (a+b+c)^2=45+2(38)=45+76=121.
\displaystyle \therefore a+b+c=\pm\sqrt{121}=\pm11.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }a+b-c=11\text{ and } a^2+b^2+c^2=89,\text{ find }ab-bc-ca.
\displaystyle \text{Answer:}
\displaystyle (a+b-c)^2 =a^2+b^2+c^2+2(ab-bc-ca).
\displaystyle \therefore 2(ab-bc-ca) =(a+b-c)^2-(a^2+b^2+c^2)
\displaystyle =11^2-89=121-89=32.
\displaystyle \therefore ab-bc-ca=16.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Expand using the identity } (a+b)^3=a^3+b^3+3ab(a+b):
\displaystyle \text{(i) }(x+3)^3\qquad \text{(ii) }(4+a)^3\qquad \text{(iii) }(6a+5b)^3
\displaystyle \text{Answer:}
\displaystyle \text{(i) }(x+3)^3 =x^3+3^3+3(x)(3)(x+3)
\displaystyle =x^3+27+9x(x+3)
\displaystyle =x^3+9x^2+27x+27.

\displaystyle \text{(ii) }(4+a)^3 =4^3+a^3+3(4)(a)(4+a)
\displaystyle =64+a^3+12a(4+a)
\displaystyle =a^3+12a^2+48a+64.

\displaystyle \text{(iii) }(6a+5b)^3 =(6a)^3+(5b)^3+3(6a)(5b)(6a+5b)
\displaystyle =216a^3+125b^3+90ab(6a+5b)
\displaystyle =216a^3+540a^2b+450ab^2+125b^3.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Expand using the identity } (a-b)^3=a^3-b^3-3ab(a-b):
\displaystyle \text{(i) }(x-4)^3\qquad \text{(ii) }(2-y)^3\qquad \text{(iii) }(4x-5y)^3
\displaystyle \text{(iv) }\left(x-\frac{1}{x}\right)^3
\displaystyle \text{Answer:}
\displaystyle \text{(i) }(x-4)^3 =x^3-4^3-3(x)(4)(x-4)
\displaystyle =x^3-64-12x(x-4)
\displaystyle =x^3-12x^2+48x-64.

\displaystyle \text{(ii) }(2-y)^3 =2^3-y^3-3(2)(y)(2-y)
\displaystyle =8-y^3-6y(2-y)
\displaystyle =8-y^3-12y+6y^2
\displaystyle =-y^3+6y^2-12y+8.

\displaystyle \text{(iii) }(4x-5y)^3 =(4x)^3-(5y)^3-3(4x)(5y)(4x-5y)
\displaystyle =64x^3-125y^3-60xy(4x-5y)
\displaystyle =64x^3-125y^3-240x^2y+300xy^2.

\displaystyle \text{(iv) }\left(x-\frac{1}{x}\right)^3
\displaystyle =x^3-\frac{1}{x^3} -3\left(x\right)\left(\frac{1}{x}\right) \left(x-\frac{1}{x}\right)
\displaystyle =x^3-\frac{1}{x^3} -3\left(x-\frac{1}{x}\right)
\displaystyle =x^3-\frac{1}{x^3}-3x+\frac{3}{x}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Evaluate the following:}
\displaystyle \text{(i) }103^3\qquad \text{(ii) }98^3\qquad \text{(iii) }402^3\qquad \text{(iv) }598^3
\displaystyle \text{Answer:}
\displaystyle \text{(i) }103^3=(100+3)^3
\displaystyle =100^3+3^3+3(100)(3)(100+3)
\displaystyle =1000000+27+900(103)
\displaystyle =1000000+27+92700
\displaystyle =1092727.

\displaystyle \text{(ii) }98^3=(100-2)^3
\displaystyle =100^3-2^3-3(100)(2)(100-2)
\displaystyle =1000000-8-600(98)
\displaystyle =1000000-8-58800
\displaystyle =941192.

\displaystyle \text{(iii) }402^3=(400+2)^3
\displaystyle =400^3+2^3+3(400)(2)(400+2)
\displaystyle =64000000+8+2400(402)
\displaystyle =64000000+8+964800
\displaystyle =64964808.

\displaystyle \text{(iv) }598^3=(600-2)^3
\displaystyle =600^3-2^3-3(600)(2)(600-2)
\displaystyle =216000000-8-3600(598)
\displaystyle =216000000-8-2152800
\displaystyle =213847192.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If } \left(a+\frac{1}{a}\right)=6,\text{ find the value of } \left(a^3+\frac{1}{a^3}\right).
\displaystyle \text{Answer:}
\displaystyle \left(a+\frac{1}{a}\right)^3 =a^3+\frac{1}{a^3}+3\left(a+\frac{1}{a}\right).
\displaystyle \therefore a^3+\frac{1}{a^3} =6^3-3(6)
\displaystyle =216-18=198.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If } \left(2a+\frac{1}{2a}\right)=4,\text{ find the value of } \left(8a^3+\frac{1}{8a^3}\right).
\displaystyle \text{Answer:}
\displaystyle \left(2a+\frac{1}{2a}\right)^3 =8a^3+\frac{1}{8a^3} +3\left(2a+\frac{1}{2a}\right).
\displaystyle \therefore 8a^3+\frac{1}{8a^3} =4^3-3(4)
\displaystyle =64-12=52.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If } \left(x-\frac{1}{x}\right)=7,\text{ find the value of } \left(x^3-\frac{1}{x^3}\right).
\displaystyle \text{Answer:}
\displaystyle \left(x-\frac{1}{x}\right)^3 =x^3-\frac{1}{x^3} -3\left(x-\frac{1}{x}\right).
\displaystyle \therefore x^3-\frac{1}{x^3} =7^3+3(7)
\displaystyle =343+21=364.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If } \left(3p-\frac{1}{3p}\right)=5,\text{ find the value of } \left(27p^3-\frac{1}{27p^3}\right).
\displaystyle \text{Answer:}
\displaystyle \left(3p-\frac{1}{3p}\right)^3 =27p^3-\frac{1}{27p^3} -3\left(3p-\frac{1}{3p}\right).
\displaystyle \therefore 27p^3-\frac{1}{27p^3} =5^3+3(5)
\displaystyle =125+15=140.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }a+b=9\text{ and } ab=20,\text{ find the value of }a^3+b^3.
\displaystyle \text{Answer:}
\displaystyle (a+b)^3=a^3+b^3+3ab(a+b).
\displaystyle \therefore a^3+b^3=(a+b)^3-3ab(a+b)
\displaystyle =9^3-3(20)(9)
\displaystyle =729-540=189.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }2x+3y=7\text{ and } xy=2,\text{ find the value of }8x^3+27y^3.
\displaystyle \text{Answer:}
\displaystyle (2x+3y)^3 =8x^3+27y^3+3(2x)(3y)(2x+3y).
\displaystyle =8x^3+27y^3+18xy(2x+3y).
\displaystyle \therefore 8x^3+27y^3 =(2x+3y)^3-18xy(2x+3y)
\displaystyle =7^3-18(2)(7)
\displaystyle =343-252=91.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }a-b=5\text{ and } ab=14,\text{ find the value of }a^3-b^3.
\displaystyle \text{Answer:}
\displaystyle (a-b)^3=a^3-b^3-3ab(a-b).
\displaystyle \therefore a^3-b^3=(a-b)^3+3ab(a-b)
\displaystyle =5^3+3(14)(5)
\displaystyle =125+210=335.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }4x-5z=2\text{ and } xz=6,\text{ find the value of }64x^3-125z^3.
\displaystyle \text{Answer:}
\displaystyle (4x-5z)^3 =64x^3-125z^3-3(4x)(5z)(4x-5z).
\displaystyle =64x^3-125z^3-60xz(4x-5z).
\displaystyle \therefore 64x^3-125z^3 =(4x-5z)^3+60xz(4x-5z)
\displaystyle =2^3+60(6)(2)
\displaystyle =8+720=728.
\displaystyle \\

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