Question paper, Set 8

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

$(3a + 7b) \times 8a = (3a)(8a) + (7b)(8a)$

$= 24a^2 + 56ab$

Answer: $24a^2 + 56ab$

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

$(2m - 7n)(4m - 3n) = (2m)(4m) - (2m)(3n) - (7n)(4m) + (7n)(3n)$

$= 8m^2 - 6mn - 28mn + 21n^2$

$= 8m^2 - 34mn + 21n^2$

Answer: $8m^2 - 34mn + 21n^2$

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

$(7m + 5n)(7m - 5n) = (7m)^2 - (5n)^2$

$= 49m^2 - 25n^2$

Answer: $49m^2 - 25n^2$

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

$\displaystyle \left(4m + \frac{1}{3}\right)\left(4m - \frac{1}{3}\right) = (4m)^2 - \left(\frac{1}{3}\right)^2$

$\displaystyle = 16m^2 - \frac{1}{9}$

Answer: $\displaystyle 16m^2 - \frac{1}{9}$

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

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

$= 64a^2 + 4b^2 + 32ab$

Answer: $64a^2 + 4b^2 + 32ab$

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

$\displaystyle \left(\frac{4x}{5} + \frac{7y}{2}\right)^2 = \left(\frac{4x}{5}\right)^2 + \left(\frac{7y}{2}\right)^2 + 2\left(\frac{4x}{5}\right)\left(\frac{7y}{2}\right)$

$\displaystyle = \frac{16x^2}{25} + \frac{49y^2}{4} + \frac{28xy}{5}$

Answer: $\displaystyle \frac{16x^2}{25} + \frac{49y^2}{4} + \frac{28xy}{5}$

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

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

$= 16a^2 + b^2 - 8ab$

Answer: $16a^2 + b^2 - 8ab$

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

$\displaystyle \left(4 + \frac{1}{x}\right)^2 + \left(4 - \frac{1}{x}\right)^2 = 2[(4)^2 + \left(\frac{1}{x}\right)^2]$

$\displaystyle = 2(16 + \frac{1}{x^2})$

$\displaystyle = 32 + \frac{2}{x^2}$

Answer: $\displaystyle 2(16 + \frac{1}{x^2}) = 32 + \frac{2}{x^2}$

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

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

$(20)^2 - (4)^2 = 32mn$

$400 - 16 = 32mn$

$\displaystyle mn = \frac{384}{32} = 12$

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

$(4m + 2n)^2 + (4m - 2n)^2 = 2[(4m)^2 + (2n)^2]$

$400 + 16 = 2(16m^2 + 4n^2)$

$\displaystyle 16m^2 + 4n^2 = \frac{416}{2} = 208$

Answer: (i) $mn = 12$ (ii) $16m^2 + 4n^2 = 208$

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

Consider $(7m + 3n)^2 - 84mn$

$= (7m + 3n)^2 - 4(7m)(3n)$

$= (7m - 3n)^2$

Answer: $(7m - 3n)^2$

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

(i) $m^2 + n^2 = (11)^2 - 2(28) = 121 - 56 = 65$

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

$= (m^2 + n^2)^2 - 2m^2n^2 = (m^2 + n^2)^2 - 2(mn)^2$

$= (65)^2 - 2(28)^2 = 4225 - 1568 = 2657$

Answer: (i) $65$ (ii) $2657$

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

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

$\displaystyle = 4m^2 + 9m^4 + \frac{16}{m^2} + 12m^3 + 24m + 16$

Answer: $\displaystyle 4m^2 + 9m^4 + \frac{16}{m^2} + 12m^3 + 24m + 16$

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

$(6a - 3b - 2c)^2 = (6a)^2 + (3b)^2 + (2c)^2 - 2(6a)(3b) + 2(3b)(2c) - 2(2c)(6a)$

$= 36a^2 + 9b^2 + 4c^2 - 36ab + 12bc - 24ca$

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

$(3a + 5b - 7c)^2 = (3a)^2 + (5b)^2 + (7c)^2 + 2(3a)(5b) - 2(5b)(7c) - 2(7c)(3a)$

$= 9a^2 + 25b^2 + 49c^2 + 30ab - 70bc - 42ca$

Answer: (i) $36a^2 + 9b^2 + 4c^2 - 36ab + 12bc - 24ca$ (ii) $9a^2 + 25b^2 + 49c^2 + 30ab - 70bc - 42ca$

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

$(q + 7)^3 = (q)^3 + (7)^3 + 3(q)(7)(q + 7)$

$= q^3 + 343 + 21q(q + 7)$

$= q^3 + 343 + 21q^2 + 147q$

Answer: $q^3 + 343 + 21q^2 + 147q$

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

$(4p - q)^3 = (4p)^3 - (q)^3 - 3(4p)(q)(4p - q)$

$= 64p^3 - q^3 - 12pq(4p - q)$

$= 64p^3 - q^3 - 48p^2q + 12pq^2$

Answer: $64p^3 - q^3 - 48p^2q + 12pq^2$

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

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

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

$= (11)^2 - 2 = 121 - 2 = 119$

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

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

$= (11)^3 - 3(11) = 1331 - 33 = 1298$

Answer: (i) $119$ (ii) $1298$

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

Here $a = z$, $b = 2$

$(z - 2)(z^2 + 2z + 4) = (z)^3 - (2)^3 = z^3 - 8$

Answer: $z^3 - 8$

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

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

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

$= (8)^3 + 3(8) = 512 + 24 = 536$

Answer: $536$

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

Here $a = 1$, $b = 9$, $c = 7$

$a + b + c = 1 + 9 + 7 = 17$

$a^2 + b^2 + c^2 - ab - bc - ca = 1 + 81 + 49 - 9 - 63 - 7 = 52$

$\text{Value} = 17 \times 52 = 884$

Answer: $884$

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

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

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

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

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