Question

factorise 2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4

Answer 1

Simplifying 2a2b2 + 2b2c2 + 2c2a2 + -1a4 + -1b4 + -1c4
 Reorder the terms: 2a2b2 + 2a2c2 + -1a4 + 2b2c2 + -1b4 + -1c4

Final result:
2a2b2 + 2a2c2 + -1a4 + 2b2c2 + -1b4 + -1c4

Answer 2

Answer:

$(b+c-a)(a+b-c)(a-b+c)(a+b+c)$

Step-by-step explanation:

2$a^{2} b^{2}$+2$b^{2} c^{2}$+2$a^{2} c^{2}$-$a^{4}-b^{4}- c^{4}$

= $- (a^{4}+ b^{4}+ c^{4}$$-2a^{2} b^{2} - 2b^{2} c^{2} -2a^{2} c^{2}$) [by reordering]

=$-[ (a^{2})^{2} + (b^{2})^{2}+ (c^{2})^{2}- 2a^{2} b^{2} -2b^{2} c^{2} +2a^{2} c^{2} -4a^{2} c^{2} ]$

= $-[ (a^{2}-b^{2}+c^{2})^{2} - 4a^{2} c^{2} ]$

[ from the formula of $(a-b+c)^{2} = a^{2}+ b^{2}+ c^{2}-2a^{2} b^{2}-2b^{2} c^{2} +2a^{2} c^{2}$]

= $-[ (a^{2}- b^{2}+ c^{2}-2ac) ( a^{2}- b^{2}+ c^{2}+2ac)]$

= $-[ (a-c)^{2}-b^{2} ][(a+c)^{2}-b^{2}]$

=$-(a-b-c)(a+b-c)(a-b+c)(a+b+c)$

=$(b+c-a)(a+b-c)(a-b+c)(a+b+c)$

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