Question paper, Set 4

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

$(5x - 6y) \times 6x = (5x)(6x) - (6y)(6x)$

$= 30x^2 - 36xy$

Answer: $30x^2 - 36xy$

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

$(5x - 9y)(5x + 4y) = (5x)(5x) + (5x)(4y) - (9y)(5x) - (9y)(4y)$

$= 25x^2 + 20xy - 45xy - 36y^2$

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

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

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

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

$= 36x^2 - 9y^2$

Answer: $36x^2 - 9y^2$

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

$\displaystyle \left(9m + \frac{1}{7}\right)\left(9m - \frac{1}{7}\right) = (9m)^2 - \left(\frac{1}{7}\right)^2$

$\displaystyle = 81m^2 - \frac{1}{49}$

Answer: $\displaystyle 81m^2 - \frac{1}{49}$

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

$(4x + 7y)^2 = (4x)^2 + (7y)^2 + 2(4x)(7y)$

$= 16x^2 + 49y^2 + 56xy$

Answer: $16x^2 + 49y^2 + 56xy$

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

$\displaystyle \left(\frac{7p}{4} + \frac{3q}{5}\right)^2 = \left(\frac{7p}{4}\right)^2 + \left(\frac{3q}{5}\right)^2 + 2\left(\frac{7p}{4}\right)\left(\frac{3q}{5}\right)$

$\displaystyle = \frac{49p^2}{16} + \frac{9q^2}{25} + \frac{21pq}{10}$

Answer: $\displaystyle \frac{49p^2}{16} + \frac{9q^2}{25} + \frac{21pq}{10}$

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

$(6m - 5n)^2 = (6m)^2 + (5n)^2 - 2(6m)(5n)$

$= 36m^2 + 25n^2 - 60mn$

Answer: $36m^2 + 25n^2 - 60mn$

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

$\displaystyle \left(1 + \frac{8}{r}\right)^2 + \left(1 - \frac{8}{r}\right)^2 = 2[(1)^2 + \left(\frac{8}{r}\right)^2]$

$\displaystyle = 2(1 + \frac{64}{r^2})$

$\displaystyle = 2 + \frac{128}{r^2}$

Answer: $\displaystyle 2(1 + \frac{64}{r^2}) = 2 + \frac{128}{r^2}$

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

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

$(19)^2 - (5)^2 = 8mn$

$361 - 25 = 8mn$

$\displaystyle mn = \frac{336}{8} = 42$

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

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

$361 + 25 = 2(4m^2 + n^2)$

$\displaystyle 4m^2 + n^2 = \frac{386}{2} = 193$

Answer: (i) $mn = 42$ (ii) $4m^2 + n^2 = 193$

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

$\text{LHS} = (m - 8n)^2 + 32mn$

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

$= (m + 8n)^2 = \text{RHS}$

Answer: LHS $= (m + 8n)^2 =$ RHS

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

(i) $m^2 + n^2 = (3)^2 + 2(18) = 9 + 36 = 45$

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

$(m + n)^2 = m^2 + n^2 + 2mn = 45 + 2(18) = 81$

Answer: (i) $45$ (ii) $81$

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

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

$\displaystyle = 9r^2 + \frac{4}{r^2} + 9r^4 + 12 + 12r + 18r^3$

Answer: $\displaystyle 9r^2 + \frac{4}{r^2} + 9r^4 + 12 + 12r + 18r^3$

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

$(4p + 3q - 7r)^2 = (4p)^2 + (3q)^2 + (7r)^2 + 2(4p)(3q) - 2(3q)(7r) - 2(7r)(4p)$

$= 16p^2 + 9q^2 + 49r^2 + 24pq - 42qr - 56rp$

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

$(3p - q - 7r)^2 = (3p)^2 + (q)^2 + (7r)^2 - 2(3p)(q) + 2(q)(7r) - 2(7r)(3p)$

$= 9p^2 + q^2 + 49r^2 - 6pq + 14qr - 42rp$

Answer: (i) $16p^2 + 9q^2 + 49r^2 + 24pq - 42qr - 56rp$ (ii) $9p^2 + q^2 + 49r^2 - 6pq + 14qr - 42rp$

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

$(5q + 1)^3 = (5q)^3 + (1)^3 + 3(5q)(1)(5q + 1)$

$= 125q^3 + 1 + 15q(5q + 1)$

$= 125q^3 + 1 + 75q^2 + 15q$

Answer: $125q^3 + 1 + 75q^2 + 15q$

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

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

$= 27n^3 - 8 - 18n(3n - 2)$

$= 27n^3 - 8 - 54n^2 + 36n$

Answer: $27n^3 - 8 - 54n^2 + 36n$

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

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

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

$= (9)^2 - 2 = 81 - 2 = 79$

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

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

$= (9)^3 - 3(9) = 729 - 27 = 702$

Answer: (i) $79$ (ii) $702$

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

Here $a = 3m$, $b = 7$

$(3m - 7)(9m^2 + 21m + 49) = (3m)^3 - (7)^3 = 27m^3 - 343$

Answer: $27m^3 - 343$

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

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

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

$= (4)^3 + 3(4) = 64 + 12 = 76$

Answer: $76$

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

Here $a = 3$, $b = 6$, $c = 1$

$a + b + c = 3 + 6 + 1 = 10$

$a^2 + b^2 + c^2 - ab - bc - ca = 9 + 36 + 1 - 18 - 6 - 3 = 19$

$\text{Value} = 10 \times 19 = 190$

Answer: $190$

$\text{Take } a = 4x - 3y,\ b = 3y - 2z,\ c = 2z - 4x$

$a + b + c = 4x - 3y + 3y - 2z + 2z - 4x = 0$

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

$\therefore (4x - 3y)^3 + (3y - 2z)^3 + (2z - 4x)^3 = 3(4x - 3y)(3y - 2z)(2z - 4x)$

Answer: $3(4x - 3y)(3y - 2z)(2z - 4x)$