Question

I need the answer now

Answer 1

i) TRUE

ii) TRUE

iii) TRUE

Explanation:

`(ab+bc)(ab-bc)+(bc+ca)(bc-ca)+(ca+ab)(ca-ab)=\\\\=\underline{(ab)^2}-(bc)^2\ \,+\ \,(bc)^2-(ca)^2\ \,+\ \,(ca)^2-\,\underline{(ab)^2}=\\\\=0+0+0=0`

`(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=\\\\=a^3\,\underline{\,+ab^2}\ \underline{\underline{+\,ac^2}}\,-a^2b\ -\,abc\ \underline{\underline{\underline{-\,a^2c}}}+a^2b+b^3+bc^2\,\underline{-\,ab^2}\ \underline{\underline{\underline{ \underline{-\,b^2c}}}}\,-\,abc+\\{}\qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \underline{\underline{ \underline{+\,a^2c}}}\ \underline{\underline{\underline{\underline{+\,b^2c}}}}+c^3\,-\,abc-bc^2\ \underline{\underline{-c^2a}}=`

`=a^3\ \underline{-\,a^2b}\ \underline{\underline{-\,abc}}\ \underline{+\,a^2b}+b^3\ \underline{\underline{\underline{+\,bc^2}}}\ \underline{\underline{-\,abc}}+c^3\ \underline{\underline{-\,abc}}\ \underline{\underline{\underline{-\,bc^2}}}=\\\\=a^3+b^3+c^3-3abc`

`(p-q)(p^2+pq+q^2)=\\\\=p^3\ \underline{+\,p^2q}\ \underline{\underline{+\,pq^2}}\ \underline{-\,p^2q}\ \underline{\underline{-\,pq^2}}-q^3=\\\\=p^3-q^3`

Answer 2

Answer: see proofs below

Explanation:

7i) (ab + bc)(ab - bc) = a²b² - ab²c + ab²c - b²c² → a²b² - b²c²

+ (bc + ca)(bc - ca) = b²c² - abc² +abc² - a²c² → b²c² - a²c²

+ (ca + ab)(ca - ab) = a²c² - a²bc + a²bc - a²b² → a²c² - a²b²

= a²b² + 0 + 0 - a²b²

= 0

7ii) a(a² + b² + c² - ab - bc - ca) = a³ + ab² + ac² - a²b - abc - a²c

+ b(a² + b² + c² - ab - bc - ca) = b³ - ab² + bc² + a²b - abc - b²c

+ c(a² + b² + c² - ab - bc - ca) = c³ - ac² - bc² - abc + a²c + b²c

= a³ + b³ + c³ + 0 + 0 + 0 + 0 - 3abc + 0 + 0

= a³ + b³ + c³ - 3abc

7iii) p(p² + pq + q²) = p³ + p²q + pq²

-q(p² + pq + q²) = - p²q - pq² - q³

= p³ + 0 + 0 - q³

= p³ - q³