Question

3 Verify the identity (a + b) (a – b) ≡ a
2 – b2
geometrically by taking
(i) a = 3 units, b = 2 units

Answer 1

$\huge\mathbb{\underline{\underline{QUESTION:-}}}$

To verify ':-

$\sf (a+b)(a-b)=a{}^{2}-b{}^{2}$

$\huge\mathbb{\blue{\underline{ANSWER:-}}}$

$\large\bf\underline{\underline{By \:Geometrically}}$

$\sf\leadsto (a+b)(a-b)\\ \sf\leadsto a{}^{2}-ab+ab-b{}^{2}\\ \sf\leadsto a{}^{2}-\cancel ab+\cancel ab -b{}^{2}\\ \sf\leadsto a{}^{2}-b{}^{2}=(a+b)(a-b)$

$\bf\underline{\underline{By \:taking \:value\: a=3\: and \:b=2}}$

$\sf \leadsto a{}^{2}-b{}^{2}=(a+b)(a-b)\\ \sf\leadsto (3){}^{2}-(2){}^{2}=(3+2)(3-2)\\ \sf\leadsto 9-4= 5×1\\ \sf\leadsto 5=5\\ \sf L.H.S=R.H.S$

$\large\bf\underline{\underline{Some\: derivation\: beginnings:-}}$

$\sf 1)(a+b){}^{2}=(a+b)(a+b)\\ \sf 2) (a-b){}^{2}=(a-b)(a-b)\\ \sf 3)(a+b+c){}^{2}=(a+b+c)(a+b+c)\\ \sf 4) a{}^{2}-b{}^{2}=(a+b)(a-b)\\ \sf 5) (a+b){}^{3}=(a+b)(a+b){}^{2}\\ \sf 6)(a-b){}^{3}=(a-b)(a-b){}^{2}\\ \sf 7)a{}^{3}+b{}^{3}=(a+b){}^{3}-3ab(a+b)\\ \sf 8) a{}^{3}-b{}^{3} =(a-b){}^{3}-3ab(a+b)$

$\large\boxed{\red{\sf{Hope\: it's \:help !!}}}$