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2026-01-01
Modified
2026-02-28
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<p>Proof of algebraic identities helps us in knowing why the identities are true and how it came to be like that. Let use try to prove the algebraic identities.</p>
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<p>Proof of algebraic identities helps us in knowing why the identities are true and how it came to be like that. Let use try to prove the algebraic identities.</p>
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<p><strong>1. Proof of \(\pmb{(a + b)^2 = a^2 + 2ab + b^2}\)</strong></p>
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<p><strong>1. Proof of \(\pmb{(a + b)^2 = a^2 + 2ab + b^2}\)</strong></p>
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<p>\((a + b)^2 = (a + b)(a + b) \\[1em] \text{Expand the equation,}\\[1em] (a + b)^2= a(a + b) + b(a + b) \\[1em] (a + b)^2= a^2 + ab + ba + b^2 \\[1em] \text{Since} \ ab=ba,\\[1em] (a + b)^2= a^2 + 2ab + b^2\)</p>
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<p>\((a + b)^2 = (a + b)(a + b) \\[1em] \text{Expand the equation,}\\[1em] (a + b)^2= a(a + b) + b(a + b) \\[1em] (a + b)^2= a^2 + ab + ba + b^2 \\[1em] \text{Since} \ ab=ba,\\[1em] (a + b)^2= a^2 + 2ab + b^2\)</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<p><strong>2. Proof of \(\pmb{(a - b)^2 = a^2 - 2ab + b^2}\)</strong></p>
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<p><strong>2. Proof of \(\pmb{(a - b)^2 = a^2 - 2ab + b^2}\)</strong></p>
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<p>\((a - b)^2 = (a - b)(a - b) \\[1em] \text{Expand the equation,}\\[1em] (a - b)^2= a(a - b) - b(a - b) \\[1em] (a - b)^2= a^2 - ab - ba + b^2 \\[1em] \text{Since}\ ab=ba,\\[1em] (a - b)^2= a^2 - 2ab + b^2 \)</p>
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<p>\((a - b)^2 = (a - b)(a - b) \\[1em] \text{Expand the equation,}\\[1em] (a - b)^2= a(a - b) - b(a - b) \\[1em] (a - b)^2= a^2 - ab - ba + b^2 \\[1em] \text{Since}\ ab=ba,\\[1em] (a - b)^2= a^2 - 2ab + b^2 \)</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<p><strong>3. Proof of \(\pmb{(a + b)(a - b) = a^2 - b^2}\)</strong></p>
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<p><strong>3. Proof of \(\pmb{(a + b)(a - b) = a^2 - b^2}\)</strong></p>
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<p>\((a + b)(a - b) = a(a - b) + b(a - b) \\[1em] \text{Upon evaluating,}\\[1em] (a + b)(a - b) = a^2 - ab + ba - b^2 \\[1em] \text{Upon simplifying},\\[1em] (a + b)(a - b) = a^2 - b^2\)</p>
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<p>\((a + b)(a - b) = a(a - b) + b(a - b) \\[1em] \text{Upon evaluating,}\\[1em] (a + b)(a - b) = a^2 - ab + ba - b^2 \\[1em] \text{Upon simplifying},\\[1em] (a + b)(a - b) = a^2 - b^2\)</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<p><strong>4. Proof of \(\pmb{(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3}\)</strong></p>
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<p><strong>4. Proof of \(\pmb{(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3}\)</strong></p>
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<p>\((a + b)^3 = (a + b)(a + b)(a + b) \\[1em] \text{Upon evaluating,}\\[1em] (a + b)^3= (a^2 + 2ab + b^2)(a + b) \\[1em] (a + b)^3= a^3 + a^2b + 2a^2b + 2ab^2 + ab^2 + b^3 \\[1em] \text{Upon simplifying,}\\[1em] (a + b)^3= a^3 + 3a^2b + 3ab^2 + b^3 \)</p>
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<p>\((a + b)^3 = (a + b)(a + b)(a + b) \\[1em] \text{Upon evaluating,}\\[1em] (a + b)^3= (a^2 + 2ab + b^2)(a + b) \\[1em] (a + b)^3= a^3 + a^2b + 2a^2b + 2ab^2 + ab^2 + b^3 \\[1em] \text{Upon simplifying,}\\[1em] (a + b)^3= a^3 + 3a^2b + 3ab^2 + b^3 \)</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<p><strong>5. Proof of \(\pmb{(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3}\)</strong></p>
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<p><strong>5. Proof of \(\pmb{(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3}\)</strong></p>
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<p>\( (a - b)^3 = (a - b)(a - b)(a - b) \\[1em] \text{Upon evaluating,}\\[1em] (a - b)^3 = (a^2 - 2ab + b^2)(a - b) \\[1em] (a - b)^3 = a^3 - a^2b - 2a^2b + 2ab^2 + ab^2 - b^3 \\[1em] \text{Upon simplifying,}\\[1em] (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \)</p>
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<p>\( (a - b)^3 = (a - b)(a - b)(a - b) \\[1em] \text{Upon evaluating,}\\[1em] (a - b)^3 = (a^2 - 2ab + b^2)(a - b) \\[1em] (a - b)^3 = a^3 - a^2b - 2a^2b + 2ab^2 + ab^2 - b^3 \\[1em] \text{Upon simplifying,}\\[1em] (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \)</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<p><strong>6. Proof of \(\pmb{a^3 + b^3 = (a + b)(a^2 - ab + b^2)}\)</strong></p>
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<p><strong>6. Proof of \(\pmb{a^3 + b^3 = (a + b)(a^2 - ab + b^2)}\)</strong></p>
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<p>\((a + b)(a^2 - ab + b^2) = a(a^2 - ab + b^2) + b(a^2 - ab + b^2) \\[1em] \text{Upon evaluating,}\\[1em] (a + b)(a^2 - ab + b^2)= a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3 \\[1em] \text{Upon simplifying,}\\[1em] (a + b)(a^2 - ab + b^2)= a^3 + b^3 \)</p>
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<p>\((a + b)(a^2 - ab + b^2) = a(a^2 - ab + b^2) + b(a^2 - ab + b^2) \\[1em] \text{Upon evaluating,}\\[1em] (a + b)(a^2 - ab + b^2)= a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3 \\[1em] \text{Upon simplifying,}\\[1em] (a + b)(a^2 - ab + b^2)= a^3 + b^3 \)</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<p><strong>7. Proof of \(\pmb{a^3 - b^3 = (a - b)(a^2 + ab + b^2)}\)</strong></p>
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<p><strong>7. Proof of \(\pmb{a^3 - b^3 = (a - b)(a^2 + ab + b^2)}\)</strong></p>
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<p>\((a - b)(a^2 + ab + b^2) = a(a^2 + ab + b^2) - b(a^2 + ab + b^2) \\[1em] \text{Upon evaluating,}\\[1em] (a - b)(a^2 + ab + b^2)= a^3 + a^2b + ab^2 - a^2b - ab^2 - b^3 \\[1em] \text{Upon simplifying,}\\[1em] (a - b)(a^2 + ab + b^2)= a^3 - b^3 \)</p>
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<p>\((a - b)(a^2 + ab + b^2) = a(a^2 + ab + b^2) - b(a^2 + ab + b^2) \\[1em] \text{Upon evaluating,}\\[1em] (a - b)(a^2 + ab + b^2)= a^3 + a^2b + ab^2 - a^2b - ab^2 - b^3 \\[1em] \text{Upon simplifying,}\\[1em] (a - b)(a^2 + ab + b^2)= a^3 - b^3 \)</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>