Factorise : ab(c^2 + d^2) - a^2cd - b^2cd

ab(c 2 \+ d 2 ) \- a 2 cd \- b 2 cd = abc 2 \+ abd 2 \- a 2 cd \- b 2 cd = (abc 2 \- a 2 cd) \+ (abd 2 \- b 2 cd) = ac(bc \- ad) \+ bd(ad \- bc) = ac(bc \- ad) \- bd(bc \- ad) = (bc \- ad)(ac \- bd) Hence,ab(c 2 \+ d 2 ) \- a 2 cd \- b 2 cd = (bc \- ad)(ac \- bd)

Mathematics

Factorise :
ab(c² + d²) - a²cd - b²cd

Factorisation

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Answer

ab(c² + d²) - a²cd - b²cd

= abc² + abd² - a²cd - b²cd

= (abc² - a²cd) + (abd² - b²cd)

= ac(bc - ad) + bd(ad - bc)

= ac(bc - ad) - bd(bc - ad)

= (bc - ad)(ac - bd)

Hence,ab(c² + d²) - a²cd - b²cd = (bc - ad)(ac - bd)

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Mathematics

Factorise :
ab(c² + d²) - a²cd - b²cd

Factorisation

Answer

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Mathematics

Factorise :ab(c2 + d2) - a2cd - b2cd

Factorisation

Answer

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ab(c² + d²) - a²cd - b²cd

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(a² - b²) c + (b² - c²)a

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Factorise :
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Factorise :
ax + 2bx + 3cx - 3a - 6b - 9c