Question

Evaluate :

(i) $(2x-\frac{3}{5})(2x+\frac{3}{5})$

(ii) $(\frac{4}{7}a+\frac{3}{4}b)(\frac{4}{7}a-\frac{3}{4}b)$

(iii) $(6-5xy)(6+5xy)$

(iv) $(2a+\frac{1}{2a})(2a-\frac{1}{2a})$

(v) $(4x^2-5y^2)(4x^2+5y^2)$

(vi) $(1.6x+0.7y)(1.6x-0.7y)$

(vii) $(m+3)(m-3)(m^2+9)$

(viii) $(3x+4y)(3x-4y)(9x^2+16y^2)$

(ix) $(a+bc)(a-bc)(a^2+b^2{c}^2)$

(x) $203\times 197$

(xi) $20.8\times 19.2$

Answer 1

(i) $(2x-\frac{3}{5})(2x+\frac{3}{5})=(2x{)}^2-{\left(\frac{3}{5}\right)}^2[\because (a-b)(a+b)=a^2{b}^2]=4x^2-\frac{9}{25}$

(ii) $(\frac{4}{7}a+\frac{3}{4}b)(\frac{4}{7}a-\frac{3}{4}b)={\left(\frac{4}{7}a\right)}^2-{\left(\frac{3}{4}b\right)}^2[\because (a-b\left)\right(a+b)=a^2{b}^2]=\frac{16}{49}{a}^2-\frac{9}{16}{b}^2$

(iii) $(6-5xy)(6+5xy)=(6{)}^2-(5xy{)}^2=36-25x^2{y}^2[\because (a-b\left)\right(a+b)=a^2{b}^2]$

(iv) $(2a+\frac{1}{2a})(2a-\frac{1}{2a})=(2a{)}^2-{\left(\frac{1}{2a}\right)}^2[\because (a-b)(a+b)=a^2{b}^2]=4a^2-\frac{1}{4a^2}$

(v) $(4x^2-5y^2)(4x^2+5y^2)=(4x^2{)}^2-(5y^2{)}^2=16x^4-25y^4[\because (a-b\left)\right(a+b)=a^2-b^2]$

(vi) $(1.6x+0.7y)(1.6x-0.7y)=(1.6x{)}^2-(0.7y{)}^2[\because (a-b\left)\right(a+b)=a^2-b^2]=2.56x^2-0.49y^2$

(vii) $(m+3)(3-3)(m^2+9)=(m{)}^2-(3{)}^2(m^2+9)[\because (a-b\left)\right(a+b)=a^2-b^2]=(m^2-9)(m^2+9)=(m^2{)}^2-9^2=m^4-81$

(viii) $(3x+4y)(3x-4y)(9x^2+16y^2)=\left[\right(3x{)}^2-\left(4y^2\right)\left]\right(9x^2+16y^2\left)\right[\because (a-b)(a+b)=a^2-b^2]=(9x^2-16y^2\left)\right(9x^2+16y^2)=(9x^2{)}^2-\left(16y^2{)}^2\right[\because (a-b)(a+b)=a^2-b^2]=81x^4-256y^4$

(ix) $(a+bc)(a-bc)(a^2+b^2{c}^2)=[a^2-(bc{)}^2\left]\right(a^2+b^2{c}^2\left)\right[\because (a-b)(a+b)=a^2-b^2]=(a^2-b^2{c}^2\left)\right(a^2+b^2{c}^2)=(a^2{)}^2-\left(b^2{c}^2{)}^2\right[\because (a-b)(a+b)=a^2-b^2]=a^4-b^4{c}^4$

(x) $203\times 197=(200+3)(200-3)=(200{)}^2-(3{)}^2=40000-9[\because (a-b\left)\right(a+b)=a^2-b^2]=39991$

(xi) $20.8\times 19.2=(20+.8)(20-.8)=(20{)}^2-(.8{)}^2[\because (a-b\left)\right(a+b)=a^2-b^2]=400-.64=399.36$