Question paper, Set 10

$a(b - c) = ab - ac$

$5p(7p - 8q) = (5p)(7p) - (5p)(8q)$

$= 35p^2 - 40pq$

Answer: $35p^2 - 40pq$

$(a - b)(c - d) = ac - ad - bc + bd$

$(3m - n)(m - 6n) = (3m)(m) - (3m)(6n) - (n)(m) + (n)(6n)$

$= 3m^2 - 18mn - mn + 6n^2$

$= 3m^2 - 19mn + 6n^2$

Answer: $3m^2 - 19mn + 6n^2$

$(a + b)(a - b) = a^2 - b^2$

$(5x + 6y)(5x - 6y) = (5x)^2 - (6y)^2$

$= 25x^2 - 36y^2$

Answer: $25x^2 - 36y^2$

$(a + b)(a - b) = a^2 - b^2$

$\displaystyle \left(5m + \frac{1}{8}\right)\left(5m - \frac{1}{8}\right) = (5m)^2 - \left(\frac{1}{8}\right)^2$

$\displaystyle = 25m^2 - \frac{1}{64}$

Answer: $\displaystyle 25m^2 - \frac{1}{64}$

$(a + b)^2 = a^2 + b^2 + 2ab$

$(6p + q)^2 = (6p)^2 + (q)^2 + 2(6p)(q)$

$= 36p^2 + q^2 + 12pq$

Answer: $36p^2 + q^2 + 12pq$

$(a + b)^2 = a^2 + b^2 + 2ab$

$\displaystyle \left(\frac{6m}{5} + \frac{3n}{5}\right)^2 = \left(\frac{6m}{5}\right)^2 + \left(\frac{3n}{5}\right)^2 + 2\left(\frac{6m}{5}\right)\left(\frac{3n}{5}\right)$

$\displaystyle = \frac{36m^2}{25} + \frac{9n^2}{25} + \frac{36mn}{25}$

Answer: $\displaystyle \frac{36m^2}{25} + \frac{9n^2}{25} + \frac{36mn}{25}$

$(a - b)^2 = a^2 + b^2 - 2ab$

$(7a - 5b)^2 = (7a)^2 + (5b)^2 - 2(7a)(5b)$

$= 49a^2 + 25b^2 - 70ab$

Answer: $49a^2 + 25b^2 - 70ab$

$(a + b)^2 + (a - b)^2 = 2(a^2 + b^2)$

$\displaystyle \left(4 + \frac{5}{p}\right)^2 + \left(4 - \frac{5}{p}\right)^2 = 2[(4)^2 + \left(\frac{5}{p}\right)^2]$

$\displaystyle = 2(16 + \frac{25}{p^2})$

$\displaystyle = 32 + \frac{50}{p^2}$

Answer: $\displaystyle 2(16 + \frac{25}{p^2}) = 32 + \frac{50}{p^2}$

(i) $(a + b)^2 - (a - b)^2 = 4ab$

$(6x + 2y)^2 - (6x - 2y)^2 = 4(6x)(2y)$

$(10)^2 - (2)^2 = 48xy$

$100 - 4 = 48xy$

$\displaystyle xy = \frac{96}{48} = 2$

(ii) $(a + b)^2 + (a - b)^2 = 2(a^2 + b^2)$

$(6x + 2y)^2 + (6x - 2y)^2 = 2[(6x)^2 + (2y)^2]$

$100 + 4 = 2(36x^2 + 4y^2)$

$\displaystyle 36x^2 + 4y^2 = \frac{104}{2} = 52$

Answer: (i) $xy = 2$ (ii) $36x^2 + 4y^2 = 52$

$(a + b)^2 = (a - b)^2 + 4ab$

$\text{LHS} = (4a - 8b)^2 + 128ab$

$= (4a - 8b)^2 + 4(4a)(8b)$

$= (4a + 8b)^2 = \text{RHS}$

Answer: LHS $= (4a + 8b)^2 =$ RHS

$a^2 + b^2 = (a + b)^2 - 2ab$

(i) $\displaystyle p^2 + \frac{1}{p^2} = (p)^2 + \left(\frac{1}{p}\right)^2$

$\displaystyle = \left(p + \frac{1}{p}\right)^2 - 2 \cdot p \cdot \frac{1}{p}$

$= (10)^2 - 2 = 100 - 2 = 98$

(ii) $(a - b)^2 = a^2 + b^2 - 2ab$

$\displaystyle \left(p - \frac{1}{p}\right)^2 = p^2 + \frac{1}{p^2} - 2 \cdot p \cdot \frac{1}{p}$

$= 98 - 2 = 96$

Answer: (i) $98$ (ii) $96$

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

$\displaystyle \left(n^2 + \frac{3}{n} + 4n\right)^2 = (n^2)^2 + \left(\frac{3}{n}\right)^2 + (4n)^2 + 2(n^2)\left(\frac{3}{n}\right) + 2\left(\frac{3}{n}\right)(4n) + 2(4n)(n^2)$

$\displaystyle = n^4 + \frac{9}{n^2} + 16n^2 + 6n + 24 + 8n^3$

Answer: $\displaystyle n^4 + \frac{9}{n^2} + 16n^2 + 6n + 24 + 8n^3$

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

$(4a + 5b - 3c)^2 = (4a)^2 + (5b)^2 + (3c)^2 + 2(4a)(5b) - 2(5b)(3c) - 2(3c)(4a)$

$= 16a^2 + 25b^2 + 9c^2 + 40ab - 30bc - 24ca$

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

$(2p - 3q - 5r)^2 = (2p)^2 + (3q)^2 + (5r)^2 - 2(2p)(3q) + 2(3q)(5r) - 2(5r)(2p)$

$= 4p^2 + 9q^2 + 25r^2 - 12pq + 30qr - 20rp$

Answer: (i) $16a^2 + 25b^2 + 9c^2 + 40ab - 30bc - 24ca$ (ii) $4p^2 + 9q^2 + 25r^2 - 12pq + 30qr - 20rp$

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

$(3t + 1)^3 = (3t)^3 + (1)^3 + 3(3t)(1)(3t + 1)$

$= 27t^3 + 1 + 9t(3t + 1)$

$= 27t^3 + 1 + 27t^2 + 9t$

Answer: $27t^3 + 1 + 27t^2 + 9t$

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

$(x - 4y)^3 = (x)^3 - (4y)^3 - 3(x)(4y)(x - 4y)$

$= x^3 - 64y^3 - 12xy(x - 4y)$

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

Answer: $x^3 - 64y^3 - 12x^2y + 48xy^2$

(i) $a^2 + b^2 = (a + b)^2 - 2ab$

$\displaystyle q^2 + \frac{1}{q^2} = (q)^2 + \left(\frac{1}{q}\right)^2$

$\displaystyle = \left(q + \frac{1}{q}\right)^2 - 2 \cdot q \cdot \frac{1}{q}$

$= (12)^2 - 2 = 144 - 2 = 142$

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

$\displaystyle q^3 + \frac{1}{q^3} = \left(q + \frac{1}{q}\right)^3 - 3 \cdot q \cdot \frac{1}{q}\left(q + \frac{1}{q}\right)$

$= (12)^3 - 3(12) = 1728 - 36 = 1692$

Answer: (i) $142$ (ii) $1692$

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Here $a = p$, $b = 4$

$(p - 4)(p^2 + 4p + 16) = (p)^3 - (4)^3 = p^3 - 64$

Answer: $p^3 - 64$

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

Here $a = x$, $\displaystyle b = \frac{1}{x}$, so $\displaystyle ab = x \cdot \frac{1}{x} = 1$

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

$= (11)^3 + 3(11) = 1331 + 33 = 1364$

Answer: $1364$

$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$

Here $a = 7$, $b = 8$, $c = 5$

$a + b + c = 7 + 8 + 5 = 20$

$a^2 + b^2 + c^2 - ab - bc - ca = 49 + 64 + 25 - 56 - 40 - 35 = 7$

$\text{Value} = 20 \times 7 = 140$

Answer: $140$

$\text{Take } a = 6x - 2y,\ b = 2y - 5z,\ c = 5z - 6x$

$a + b + c = 6x - 2y + 2y - 5z + 5z - 6x = 0$

If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

$\therefore (6x - 2y)^3 + (2y - 5z)^3 + (5z - 6x)^3 = 3(6x - 2y)(2y - 5z)(5z - 6x)$

Answer: $3(6x - 2y)(2y - 5z)(5z - 6x)$