Question 1: Expand:

$i) (2a+3b+4c)^2$

$= (4a^2 +9b^(2 )+16c^2) +2 (6ab+12bc+8ac)$

$= 4a^(2 )+9b^2+16c^2+12ab+24bc+16ac$

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$ii) (a+2b-5c)^2$

$= (a^2+4b^2+25c^2) +2 (2ab -10bc-5ac)$

$= a^2+4b^2+25c^2+4ab-20bc-10ac$

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$iii) (3x +2y-z)^2$

$= (9x^2+4y^2+z^2 ) +2 (6xy-2yz-6xz)$

$= 9x^2+4y^2+z^2+12xy-4yz-12xz$

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$iv) (3x-2y-1)^2$

$= (9x^2+4y^2+1)+ 2 (-6xy+2y-3x)$

$= 9x^2+4y^2+1-12xy+4y-6x$

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$v) (\frac{a}{2}-\frac{b}{3}+\frac{c}{4})^2$

$= (\frac{a^2}{4}+\frac{b^2}{9}+\frac{c^2}{16})+2(\frac{-ab}{6}-\frac{bc}{12}+\frac{ac}{8})$

$= (\frac{a^2}{4}+\frac{b^2}{9}+\frac{c^2}{16})-\frac{ab}{3}-\frac{bc}{6}+\frac{ac}{4}$

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$vi) (x-\frac{2}{x}+3)^2$

$= x^2 +\frac{4}{x^2} +9-2\times (-2-\frac{6}{x}+3x)$

$= x^2+\frac{4}{x^2} +9+4+\frac{12}{x}-6x$

$= x^2+\frac{4}{x^2} +\frac{12}{x}-6x+13$

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Question 2: If $(a+b+c)=10$   and $ab+bc+ac = 31$ Find $(a^2+b^2+c^2 )$

$(a+b+c)^2=(a^2+b^2+c^2 )+2(ab+bc+ca)$

$(a^2+b^2+c^2 )=10^2-2\times 3=100-62=38$

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Question 3: If $((a+b+c)=9$ and $(a^2+b^2+c^2 )=29$ find $(ab+bc+ac)$

$(a+b+c)^2=(a^2+b^2+c^2 )+2(ab+bc+ac)$

$(ab+bc+ac)=\frac{1}{2} (9^2-29)=\frac{(81-29)}{2}=26$

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Question 4: If  $(a^2+b^2+c^2 )=45$ and  $(ab+bc+ac)=38$ find$(a+b+c)$

$(a+b+c)^2=(a^2+b^2+c^2 )+2(ab+bc+ac)$

$(a+b+c)^2=45+2\times 38=121$

$(a+b+c)=11$

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Question 5: If $(a+b-c)= 11$  and $(a^2+b^2+c^2 )=89$  find $(ab-bc-ca)$

$(a+b-c)^2=(a^2+b^2+c^2 )+2(ab-bc-ac)$

or $(ab-bc-ca)=\frac{1}{2} (81-11^2 )=-20$

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Question 6: Expanding using Formula    $(a+b)^3=a^3+b^3+3ab(a+b)$

$i) (x+3)^3= x^3+27+9x(x+3)$

$= x^3+9x^2+27x+27$

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$ii) (4+a)^3=64+a^3+3\times4a (4+a)$

$= a^3+12a^2+12a+64$

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$iii) (6a+5b)^3= (6a)^3+(5b)^3+3(6a)(5b)(6a+5b)$

$=216a^3+125b^3+540a^2 b+450ab^2$

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Question 7: Expand Unity Formula $(a-b)^3=a^3-b^3-3ab(a-b)$

(i) $(x-4)^3=x^3-64-12x(x-4)$

$= x^3-12x^2+48x-64$

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(ii) $(2-y)^3=8-y^3-4y(2-y)$

$= -y^3+4y-8y+8$

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(iii) $(4x-5y)^3=64 x^3-125y^3-60xy(4x-5y)$

$=64x^3-125y^3-240x^2 y+300 xy^2$

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Question 8: Evaluate the following:

$a) \ \ 103^3=(100+3)^3=1000000+27+900(100+3)$

$= 1000000+27+92700$

$= 109272.7$

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$b) \ \ 98^3=(100-2)^3=1000000-8-600(100-2)$

$= 1000000-8-58800$

$=941192$

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$c) \ \ 402^3=(400+2)^3=64000000+8+2400(400+2)$

$= 64964808$

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$d) \ \ 598^3=(600-2)^(3 )=216000000-8-3600(600-2)$

$= 216000000-8-2152800$

$= 213847192$

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Question 9: If $(a+\frac{1}{a})=6,$   find  the value of$(a^3+\frac{1}{a^3})$

$(a+\frac{1}{a})^3= a^3+ \frac{1}{a^3} +3 (a+\frac{1}{a})$

$or\ \ a^3+\frac{1}{a^3} =6^3-36=198$

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Question 10: If $(2a +\frac{1}{2a})=4$  Find the value of  $(8a^3+ \frac{1}{8a^3} )$

$(2a +\frac{1}{2a})^3=8a^3+ \frac{1}{8a^3}+3(2a +\frac{1}{2a})$

$or \ \ 8a^3+ \frac{1}{8a^3}=4^3-3\times 4=64-12=52$

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Question 11: If $(x- \frac{1}{x})=7$    Find the value of $(x^3- \frac{1}{x^3} )$

$(x- \frac{1}{x})^3=x^3- \frac{1}{x^3} -3(x- \frac{1}{x})$

$\ \ or x^3- \frac{1}{x^3} =7^3+3\times 7=343 +21=364$

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Question 12: If $(3p- \frac{1}{3p})=5,$    Find the value of $(27p^3- \frac{1}{27 p^3})$

$(3p- \frac{1}{3p})^3= (27 p^3-\frac{1}{27 p^3} )-3 (3p- \frac{1}{3p})$

$or \ \ (27p^3-\frac{1}{27 p^3})=5^3+3\times 5=125 + 15 =140$

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Question 13: If $a + b = 9$   and $ab=20$   find the value of $a^3+b^3$

$(a+b)^3 = a^3+b^3+3ab(a+b)$

$or \ \ a^3+ b^3= 9^3-3\times 20\times 9=189$

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Question 14: If$(2x + 3y) = 7$   and $xy=2$   find the value of $(8x^3+27y^3 )$

$(2x+3y)^3= 8x^3+27y^3+18 xy (2x+3y)$

$8x^3+27y^3=7^3-18\times 2\times 7=91$

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Question 15:  If $(a-b)=5$   and $ab=14$   Find the value of $(a^3-b^3)$

$(a-b)^3 = (a^3-b^3 )-3ab(a-b)$

$a^3-b^3=5^3-3\times 5\times 14=-85$

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Question 16:  If $(4x-5z)=2$   and $xz=6$   find the value of $(64x^3 -125z^3)$

$(4x-5z)^3=64x^3 -125x^3 -60xz (4x-5z)$
$64^3-125^3=23+60 \times 6 \times 2 = 728$
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