Question paper, Set 7

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

$(3m + 7n) \times 7m = (3m)(7m) + (7n)(7m)$

$= 21m^2 + 49mn$

Answer: $21m^2 + 49mn$

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

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

$= 20x^2 - 45xy - 4xy + 9y^2$

$= 20x^2 - 49xy + 9y^2$

Answer: $20x^2 - 49xy + 9y^2$

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

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

$= 4m^2 - 64n^2$

Answer: $4m^2 - 64n^2$

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

$(2a^2 + 6b^5)(2a^2 - 6b^5) = (2a^2)^2 - (6b^5)^2$

Using $(ab)^n = a^n \times b^n$: $(2a^2)^2 = 4a^4$ and $(6b^5)^2 = 36b^{10}$

$= 4a^4 - 36b^{10}$

Answer: $4a^4 - 36b^{10}$

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

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

$= 4m^2 + 25n^2 + 20mn$

Answer: $4m^2 + 25n^2 + 20mn$

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

$\displaystyle \left(3z + \frac{1}{5}\right)^2 = (3z)^2 + \left(\frac{1}{5}\right)^2 + 2(3z)\left(\frac{1}{5}\right)$

$\displaystyle = 9z^2 + \frac{1}{25} + \frac{6z}{5}$

Answer: $\displaystyle 9z^2 + \frac{1}{25} + \frac{6z}{5}$

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

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

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

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

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

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

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

$= 2x^2 + 8y^2$

Answer: $2(x^2 + 4y^2) = 2x^2 + 8y^2$

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

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

$(17)^2 - (7)^2 = 16ab$

$289 - 49 = 16ab$

$\displaystyle ab = \frac{240}{16} = 15$

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

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

$289 + 49 = 2(16a^2 + b^2)$

$\displaystyle 16a^2 + b^2 = \frac{338}{2} = 169$

Answer: (i) $ab = 15$ (ii) $16a^2 + b^2 = 169$

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

$\text{LHS} = (3m - n)^2 + 12mn$

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

$= (3m + n)^2 = \text{RHS}$

Answer: LHS $= (3m + n)^2 =$ RHS

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

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

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

$= (5)^2 + 2 = 25 + 2 = 27$

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

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

$= (27)^2 - 2 = 729 - 2 = 727$

Answer: (i) $27$ (ii) $727$

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

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

$= 4p^2 + 16q^2 + 9r^2 + 16pq + 24qr + 12rp$

Answer: $4p^2 + 16q^2 + 9r^2 + 16pq + 24qr + 12rp$

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

$(7p + 5q - 4r)^2 = (7p)^2 + (5q)^2 + (4r)^2 + 2(7p)(5q) - 2(5q)(4r) - 2(4r)(7p)$

$= 49p^2 + 25q^2 + 16r^2 + 70pq - 40qr - 56rp$

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

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

$= 4p^2 + 49q^2 + 25r^2 - 28pq - 70qr + 20rp$

Answer: (i) $49p^2 + 25q^2 + 16r^2 + 70pq - 40qr - 56rp$ (ii) $4p^2 + 49q^2 + 25r^2 - 28pq - 70qr + 20rp$

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

$(t + 5)^3 = (t)^3 + (5)^3 + 3(t)(5)(t + 5)$

$= t^3 + 125 + 15t(t + 5)$

$= t^3 + 125 + 15t^2 + 75t$

Answer: $t^3 + 125 + 15t^2 + 75t$

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

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

$= m^3 - 27n^3 - 9mn(m - 3n)$

$= m^3 - 27n^3 - 9m^2n + 27mn^2$

Answer: $m^3 - 27n^3 - 9m^2n + 27mn^2$

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

$a^2 + b^2 = (9)^2 - 2(18) = 81 - 36 = 45$

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

$a^3 + b^3 = (9)^3 - 3(18)(9)$

$= 729 - 486 = 243$

Check with $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$: $(9)(45 - 18) = 9 \times 27 = 243$

Answer: (i) $45$ (ii) $243$

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

Here $a = 2z$, $b = 5$

$(2z - 5)(4z^2 + 10z + 25) = (2z)^3 - (5)^3 = 8z^3 - 125$

Answer: $8z^3 - 125$

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

$p^3 - q^3 = (3)^3 + 3(54)(3)$

$= 27 + 486 = 513$

Answer: $513$

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

Here $a = 3p$, $b = q$, $c = 5$

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

$= 27p^3 + q^3 + 125 - 45pq$

Answer: $27p^3 + q^3 + 125 - 45pq$

$\text{Take } a = 10,\ b = -13,\ c = 3$

$a + b + c = 10 - 13 + 3 = 0$

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

$\therefore 10^3 + (-13)^3 + 3^3 = 3(10)(-13)(3) = -1170$

Answer: $-1170$