Question paper, Set 2

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

$7a(9a + 5b) = (7a)(9a) + (7a)(5b)$

$= 63a^2 + 35ab$

Answer: $63a^2 + 35ab$

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

$(3x - 9y)(4x + 4y) = (3x)(4x) + (3x)(4y) - (9y)(4x) - (9y)(4y)$

$= 12x^2 + 12xy - 36xy - 36y^2$

$= 12x^2 - 24xy - 36y^2$

Answer: $12x^2 - 24xy - 36y^2$

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

$(p - 7q)(p + 7q) = (p)^2 - (7q)^2$

$= p^2 - 49q^2$

Answer: $p^2 - 49q^2$

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

$\displaystyle \left(9n + \frac{1}{4}\right)\left(9n - \frac{1}{4}\right) = (9n)^2 - \left(\frac{1}{4}\right)^2$

$\displaystyle = 81n^2 - \frac{1}{16}$

Answer: $\displaystyle 81n^2 - \frac{1}{16}$

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

$(6m + 4n)^2 = (6m)^2 + (4n)^2 + 2(6m)(4n)$

$= 36m^2 + 16n^2 + 48mn$

Answer: $36m^2 + 16n^2 + 48mn$

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

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

$\displaystyle = \frac{p^2}{9} + \frac{36q^2}{25} + \frac{4pq}{5}$

Answer: $\displaystyle \frac{p^2}{9} + \frac{36q^2}{25} + \frac{4pq}{5}$

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

$(8m - 4n)^2 = (8m)^2 + (4n)^2 - 2(8m)(4n)$

$= 64m^2 + 16n^2 - 64mn$

Answer: $64m^2 + 16n^2 - 64mn$

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

$\displaystyle \left(8 + \frac{3}{p}\right)^2 + \left(8 - \frac{3}{p}\right)^2 = 2[(8)^2 + \left(\frac{3}{p}\right)^2]$

$\displaystyle = 2(64 + \frac{9}{p^2})$

$\displaystyle = 128 + \frac{18}{p^2}$

Answer: $\displaystyle 2(64 + \frac{9}{p^2}) = 128 + \frac{18}{p^2}$

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

$(m + 6n)^2 - (m - 6n)^2 = 4(m)(6n)$

$(13)^2 - (1)^2 = 24mn$

$169 - 1 = 24mn$

$\displaystyle mn = \frac{168}{24} = 7$

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

$(m + 6n)^2 + (m - 6n)^2 = 2[(m)^2 + (6n)^2]$

$169 + 1 = 2(m^2 + 36n^2)$

$\displaystyle m^2 + 36n^2 = \frac{170}{2} = 85$

Answer: (i) $mn = 7$ (ii) $m^2 + 36n^2 = 85$

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

Consider $(6x + 5y)^2 - 120xy$

$= (6x + 5y)^2 - 4(6x)(5y)$

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

Answer: $(6x - 5y)^2$

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

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

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

$= (2)^2 + 2 = 4 + 2 = 6$

(ii) $\displaystyle q^4 + \frac{1}{q^4} = (q^2)^2 + \left(\frac{1}{q^2}\right)^2$

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

$= (6)^2 - 2 = 36 - 2 = 34$

Answer: (i) $6$ (ii) $34$

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

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

$\displaystyle = 9y^4 + \frac{9}{y^2} + y^2 + 18y + 6 + 6y^3$

Answer: $\displaystyle 9y^4 + \frac{9}{y^2} + y^2 + 18y + 6 + 6y^3$

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

$(7p - 5q + r)^2 = (7p)^2 + (5q)^2 + (r)^2 - 2(7p)(5q) - 2(5q)(r) + 2(r)(7p)$

$= 49p^2 + 25q^2 + r^2 - 70pq - 10qr + 14rp$

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

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

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

Answer: (i) $49p^2 + 25q^2 + r^2 - 70pq - 10qr + 14rp$ (ii) $25a^2 + b^2 + 4c^2 - 10ab + 4bc - 20ca$

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

$(6x + 1)^3 = (6x)^3 + (1)^3 + 3(6x)(1)(6x + 1)$

$= 216x^3 + 1 + 18x(6x + 1)$

$= 216x^3 + 1 + 108x^2 + 18x$

Answer: $216x^3 + 1 + 108x^2 + 18x$

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

$(3m - 1)^3 = (3m)^3 - (1)^3 - 3(3m)(1)(3m - 1)$

$= 27m^3 - 1 - 9m(3m - 1)$

$= 27m^3 - 1 - 27m^2 + 9m$

Answer: $27m^3 - 1 - 27m^2 + 9m$

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

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

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

$= (5)^2 - 2 = 25 - 2 = 23$

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

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

$= (5)^3 - 3(5) = 125 - 15 = 110$

Answer: (i) $23$ (ii) $110$

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

Here $a = 2r$, $b = 4$

$(2r - 4)(4r^2 + 8r + 16) = (2r)^3 - (4)^3 = 8r^3 - 64$

Answer: $8r^3 - 64$

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

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

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

$= (3)^3 + 3(3) = 27 + 9 = 36$

Answer: $36$

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

Here $a = 2$, $b = 9$, $c = 8$

$a + b + c = 2 + 9 + 8 = 19$

$a^2 + b^2 + c^2 - ab - bc - ca = 4 + 81 + 64 - 18 - 72 - 16 = 43$

$\text{Value} = 19 \times 43 = 817$

Answer: $817$

$\text{Take } a = 2m - 4n,\ b = 4n - p,\ c = p - 2m$

$a + b + c = 2m - 4n + 4n - p + p - 2m = 0$

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

$\therefore (2m - 4n)^3 + (4n - p)^3 + (p - 2m)^3 = 3(2m - 4n)(4n - p)(p - 2m)$

Answer: $3(2m - 4n)(4n - p)(p - 2m)$